Problems on Expanding of (a ± b)^3 and its Corollaries

Problems on Expanding of (a ± b)^3 and its Corollaries | Expansion of (a ± b)^3 and its Corollaries Questions and Answers

Problems on Expanding of (a ± b)^3 and its Corollaries are provided on this page. Students who are searching for the concept Expanding of (a ± b)^3 can get them here. Solve the given problems to know what you have learned from the Expansion of Powers of Binomials and Trinomials chapter. This helps you to overcome the difficulties in this chapter and secure good marks in the exams. Hence we suggest the students practice the given problems before they go for the exam.

Do Refer:

Questions on Expanding of (a ± b)^3 and its Corollaries

Example 1.
Expand (3 + x)³ using the formula (a+b)³
Solution:
Given the binomial expression (3 + x)³
We know that,
(a+b)³ = a³ + 3a²b + 3ab² + b³
Substitute the given values in the standard formula.
(3 + x)³ = 3³ + 3 × 3²x + 3 × 3x² + x³
(3 + x)³ = 27 + 27x + 9x² + x³
Thus the expansion of (3 + x)³ = 27 + 27x + 9x² + x³

Example 2.
If a + b = 5, the cubes of the sum of two terms is 10 then find the ab.
Solution:
Given that the sum of two terms is 5.
a + b = 5
a³ + b³ = 10
By using the cubic formula (a+b)³ = a³ + 3a²b + 3ab² + b³ We can find the product of the constant terms.
(5)³ = 10 + 3a²b + 3ab²
125 = 10 + 3ab(a + b)
125 – 10 = 3ab(5)
115 = 15ab
115/15 = ab
ab = 7.66
Thus the product of the two constant terms is 7.66

Example 3.
If the sum of the two constant terms is 10 and the a³ + b³ = 12. Find the product of two numbers ab.
Solution:
Given that,
a + b = 10
a³ + b³ = 12
By using the cubic formula (a+b)³ = a³ + 3a²b + 3ab² + b³ We can find the product of the constant terms.
(a+b)³ = a³ + 3a²b + 3ab² + b³
(10)³ = 12 + 3a²b + 3ab²
1000 = 12 + 3ab(a + b)
1000 – 12 = ab(30)
988/30 = ab
ab = 32.9
Thus the product of the two terms is 32.9

Example 4.
Expand (x – 8)³ using the formula (a – b)³
Solution:
Given the expression (x- 8)³,
This is in the form of (a – b)³ = a³ – 3a²b + 3ab² – b³
Substitute the values in the given formula
(x – 8)³ = x³ – 3x²8 + 3×8² – 8³
(x – 8)³ = x³ – 24x² + 192x – 512
Thus the expansion of (x – 8)³ = x³ – 24x² + 192x – 512

Example 5.
Expand the expression (a + 6)³ using the formula (a+b)³.
Solution:
Given (a + 6)³
We know that,
(a+b)³ = a³ + 3a²b + 3ab² + b³
Substitute the given values in the standard formula.
(a+6)³ = a³ + 3a²b + 3ab² + b³
(a+6)³ = a³ + 3a²6 + 3a6² + 6³
(a+6)³ = a³ + 18a² + 108a + 216
Thus the expansion of (a+6)³ is a³ + 18a² + 108a + 216

Example 6.
Find the value of (1 + 4)³ using the formula (a+b)³.
Solution:
We know that,
(a+b)³ = a³ + 3a²b + 3ab² + b³
Substitute the given values in the standard formula.
(1+4)³ = 1³ + 3(1)²(4) + 3(1)(4)² + (4)³
(1+4)³ = 1 + 12 + 48 + 64
(1+4)³ = 125
Thus the value of (1 + 4)³ is 125.

Example 7.
Find the value of the following expressions
i. (5 – 2)³
ii. (8 – 1)³
iii. (6 + 4)³
Solution:
i. (5 – 2)³
Given the expression (5 – 2)³,
This is in the form of (a – b)³ = a³ – 3a²b + 3ab² – b³
Substitute the values in the given formula
(5 – 2)³ = 5³ – 3(5)²(2) + 3(5)(2)² – (2)³
(5 – 2)³ = 125 – 150 + 60 – 8
(5 – 2)³ = 27
ii. (8 – 1)³
Given the expression (8 – 1)³,
This is in the form of (a – b)³ = a³ – 3a²b + 3ab² – b³
Substitute the values in the given formula
(8 – 1)³ = 8³ – 3(8)²(1) + 3(8)(1)² – (1)³
(5 – 2)³ = 512 – 192 + 24 – 1
(5 – 2)³ = 343
iii. (6 + 4)³
Given the expression (6 + 4)³,
We know that,
This is in the form of (a+b)³ = a³ + 3a²b + 3ab² + b³
Substitute the given values in the standard formula.
(6+4)³ = 6³ + 3(6)²(4) + 3(6)(4)² + (4)³
(6+4)³ = 216 + 432 + 144 + 64
(6+4)³ = 856
Thus the value of (6+4)³ is 856

Example 8.
Find a³ + b³ if the sum of the terms is 7 and the product of the terms is 10.
Solution:
Given,
The sum of the terms is 7
a + b = 7
the product of the terms is 10
ab = 10
a³ + b³ = ?
We know that,
(a+b)³ = a³ + 3a²b + 3ab² + b³
Substitute the given values in the standard formula.
(7)³ = a³ + 3a²b + 3ab² + b³
(7)³ = a³ + 3(10)(7) + b³
(7)³ = a³ + b³ + 210
343 – 210 = a³ + b³
a³ + b³ = 133

Example 9.
If a + b = 3 and ab = 6 then find the cubes of sum of the two constant terms.
Solution:
Given,
a + b = 3 and ab = 6
We know that,
(a+b)³ = a³ + 3a²b + 3ab² + b³
Substitute the given values in the standard formula.
3³ = a³ + 3ab(a + b) + b³
27 = a³ + 3(6)(3) + b³
27 – 54 = a³ + b³
a³ + b³ = -27
Therefore, the cubes of the sum of the two constant terms are -27.

Example 10.
If x – y = 5 and x³ – y³ = 15 then find the product of the terms.
Solution:
Given,
x – y = 5 and
x³ – y³ = 15
This is in the form of (a – b)³ = a³ – 3a²b + 3ab² – b³
Substitute the values in the given formula
(5)³ = 15 – 3a²b + 3ab²
125 – 15 = -3ab(a – b)
110 = -3ab(5)
110/-15 = ab
ab = -7.33
Thus the product of the terms is -7.33

Expansion of (x + a)(x + b)(x + c)

Expansion of (x + a)(x + b)(x + c) – Derivations, Examples | How to Expand the Product of Expression (x + a)(x + b)(x + c)?

Expansion of (x + a)(x + b)(x + c): The key to understanding the concept of Expansion of Powers of Binomials and Trinomials is through brackets. You can understand how to multiply the binomial expression with the help of brackets. In this article you can find quick and easy methods for Expansion of (x + a)(x + b)(x + c). Scroll down this page to know the derivation of Expansion of (x + a)(x + b)(x + c) with suitable examples.

Expansion of (x + a)(x + b)(x + c)

Here we will discuss the expansion of (x + a)(x + b)(x + c) with the derivation with explanation.
(x + a)(x + b)(x + c) = (x + a){(x + b)(x + c)}
(x + a)(x + b)(x + c) = x {(x + b)(x + c)} + a {(x + b)(x + c)}
(x + a)(x + b)(x + c) = x {x(x + c) + b(x + c)} + a {x(x + c) + b(x + c)}
(x + a)(x + b)(x + c) = x {x² + cx + bx + bc} + a {x² + cx + bx + bc}
(x + a)(x + b)(x + c) = x³ + cx² + bx² + bcx + ax² + acx + abx + abc
(x + a)(x + b)(x + c) = x³ + x²(a + b + c) + x(ab + bc + ac) + abc
Thus the expansion of (x + a)(x + b)(x + c) is x³ + x²(a + b + c) + x(ab + bc + ac) + abc
This can be read as cubed x + squared x {sum of the three constant terms} + x{sum of the product of the three constant terms} + product of three constant terms.

Do Refer:

Expansion of (x + a)(x + b)(x + c) Examples

Step by step explanation for each and every question is provided here.

Example 1.
Find the product of (x + 1) (x + 3) (x + 6) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x + 1) (x + 3) (x + 6)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = 1, b = 3, c = 6
Therefore
(x + 1) (x + 3) (x + 6) = x³ + (1 + 3 + 6)x² + (1 × 3 + 3 × 6 + 6 × 1)x + 1× 3× 6
(x + 1) (x + 3) (x + 6) = x³ + 10x² + 27x + 18
Thus the product of (x + 1) (x + 3) (x + 6) is x³ + 10x² + 27x + 18

Example 2.
Find the product of (x + 2) (x – 1) (x -3) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x + 2) (x – 1) (x – 3)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = 1, b = 3, c = 6
Therefore
(x + 2) (x – 1) (x – 3) = x³ + (2-1-3)x² + (2(-1)+(-1)(-3)+(-3)(2))x + 2(-1)(-3)
(x + 2) (x – 1) (x – 3) = x³ + (-2)x² + (-2+3-6)x + 6
(x + 2) (x – 1) (x – 3) = x³ – 2x² + 7x + 6
Thus the product of (x + 2) (x – 1) (x – 3) is x³ – 2x² + 7x + 6

Example 3.
Find the product of (x – 1) (x – 2) (x – 3) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x – 1) (x – 2) (x – 3)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = -1, b = -2, c = -3
Therefore
(x – 1) (x – 2) (x – 3) = x³ + (-1-2-3)x² + ((-1)(-2)+(-2)(-3)+(-3)(-1))x + (-1)(-2)(-3)
(x – 1) (x – 2) (x – 3) = x³ – 6x² + 11x – 6
Thus the product of (x – 1) (x – 2) (x – 3) is x³ – 6x² + 11x – 6

Example 4.
Find the product of (x + 3) (x +4) (x +6) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x + 3) (x + 4) (x + 6)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = 3, b = 4, c = 6
Therefore
(x + 3) (x + 4) (x + 6) = x³ + (3 + 4 + 6)x² + (3×4 + 4×6 6×3)x + 3×4×6
(x + 3) (x + 4) (x + 6) = x³ + 13x² + 54x + 72
Thus the product of (x + 3) (x + 4) (x + 6) is x³ + 13x² + 54x + 72

Example 5.
Find the product of (x – 1) (x + 2) (x + 4) using the formula (x + a)(x + b)(x + c).
Solution:
Given that
(x – 1) (x + 2) (x + 4)
We know that
(x + a)(x + b)(x + c) = x³ + (a + b + c)x² + (ab + bc + ca)x + abc
Here
a = -1, b = 3, c = 6
Therefore
(x – 1) (x + 2) (x + 4) = x³ + (-1 + 3 + 6)x² + (-1(3) + 3 × 6 + 6(-1))x + (-1) × 3 × 6
(x – 1) (x + 2) (x + 4) = x³ + 8x² + 9x – 18
Thus the product of (x – 1) (x + 2) (x + 4) is x³ + 8x² + 9x – 18

Simplification of (a ± b)(a^2 ∓ ab + b^2)

Simplification of (a ± b)(a^2 ∓ ab + b^2) – Derivation, Examples | How to Simplify the Identity (a^3 ± b^3)

It is very important for your child to know how to expand and how to simplify the binomial and trinomial expressions. Without knowing the simplifications they cannot solve the problems. On this page, you can find the methods to simplify (a ± b)(a^2 ∓ ab + b^2) expressions. Let us take some examples of simplification of (a ± b)(a^2 ∓ ab + b^2) and solve them in simple techniques.

Solving the problems in simple methods will help you to finish the exam time. So, we suggest the students practice the problems on Expansion of Powers of Binomials and Trinomials to become perfect in this chapter.

Simplification of (a ± b)(a^2 ∓ ab + b^2) Derivation

Simplification of (a + b)(a² – ab + b²):
(a + b)(a² – ab + b²) = a(a² – ab + b²) + b (a² – ab + b²)
(a + b)(a² – ab + b²) = a³ – a²b + ab² + ba² – ab² + b³
(a + b)(a² – ab + b²) = a³ + b³

Simplification of (a – b)(a² +ab + b²):
(a – b)(a² + ab + b²) = a(a² + ab + b²) – b (a² + ab + b²)
(a – b)(a² + ab + b²) = a³ + a²b + ab² – ba² – ab² – b³
(a – b)(a² + ab + b²) = a³ – b³

Read Similar Articles:

Simplifying (a ± b)(a^2 ∓ ab + b^2) Examples

The simplification of (a ± b)(a^2 ∓ ab + b^2) is shown with some examples in the below section.

Example 1.
Simply the equation (3x + y) (9x² – 3xy + y²)
Solution:
Given that
(3x + y) (9x² – 3xy + y²)
(3x + y) {(3x)² – (3y)y + y²}
(3x)³ + y³ [ therefore (a + b)(a² – ab + b² = a³ + b³]
27x³ + y³
The simplification of (3x + y) (9x² – 3xy + y²) is 27x³ + y³

Example 2.
Simply the equation (x + 1/x)(x² + 1 + 1/x²)
Solution:
Given that
(x + 1/x)(x² + 1 + 1/x²)
(x + 1/x)( x² + x × 1/x + (1/x)³)
x³ – 1/x³ [ Since (a + b)(a² – ab + b² = a³ + b³]
The simplification of (x + 1/x)(x² + 1 + 1/x²) is x³ – 1/x³

Example 3.
Simply the equation (4x + y) (16x² – 4xy + y²)
Solution:
Given that
(4x + y) (16x² – 4xy + y²)
(4x + y) {(4x)² – (4y)y + y²}
(4x)³ + y³ [ since the formula of (a + b)(a² – ab + b² = a³ + b³]
64x³ + y³
The simplification of (4x + y) (16x² – 4xy + y²) is 64x³ + y³

Example 4.
Simply the equation (5x + y) (25x² – 5xy + y²)
Solution:
Given that
(5x + y) (25x² – 5xy + y²)
(5x + y) {(5x)² – (5y)y + y²}
(5x)³ + y³ [ therefore (a + b)(a² – ab + b² = a³ + b³]
125x³ + y³
The simplification of (5x + y) (25x² – 5xy + y²) is 125x³ + y³

Example 5.
Simply the equation (6x + y) (36x² – 6xy + y²)
Solution:
Given that
(6x + y) (36x² – 6xy + y²)
(6x + y) {(6x)² – (6y)y + y²}
(6x)³ + y³ [ therefore (a + b)(a² – ab + b² = a³ + b³]
216x³ + y³
The simplification of (6x + y) (36x² – 6xy + y²) is 216x³ + y³

Expansion of (a ± b)^3

Expansion of (a ± b)^3 – Derivation, Formula, Examples | How do you Expand (a ± b)^3?

The formula (a + b)³ and (a – b)³ is used to find the cube of the binomial. The binomial theorem is an expansion of the algebraic expression. Expansion of the binomial expansion is the product of each constant term with the expression. For your convenience, we have provided the derivation of (a + b)³ and (a – b)³ along with the corollaries. Also, you can find suitable examples of Expansion of Powers of Binomials and Trinomials from this page.

Expansion of (a ± b)^3 Derivation

Expansion of (a + b)^3:
(a + b)³ can be written as (a + b)(a + b)²
(a + b)³ = (a + b)(a + b)²
(a + b)³ = (a + b)[a² + 2ab + b²]
(a + b)³ = a [a² + 2ab + b²] + b [a² + 2ab + b²]
(a + b)³ = a³ + 2a²b + ab² + ba² + 2ab² + b³
(a + b)³ = a³ + 3a²b + 3ab² + b³
(a + b)³ = a³ + 3ab(a + b) + b³
Thus the formula of (a + b)³ = a³ + 3ab(a + b) + b³
It can be read as cubed a + 3 × product of two terms (sum of the two constant terms) + cubed b

Expansion of (a – b)^3:
(a – b)³ can be written as (a – b)(a – b)²
(a – b)³ = (a – b)(a – b)²
(a – b)³ = (a – b) [a² + b² – 2ab]
(a – b)³ = a[a² + b² – 2ab] -b [a² + b² – 2ab]
(a – b)³ = a³ + ab² – 2a²b – ba² – b³ + 2ab²
(a – b)³ = a³ + 3ab² – 3a²b – b³
(a – b)³ = a³ + 3ab(a – b) – b³
Thus the formula of (a – b)³ = a³ + 3ab(a – b) – b³
It can be read as cubed a + 3 × product of two terms (difference of the two constant terms) – cubed b

Corollaries:

  1. (a + b)³ = a³ + 3ab(a + b) + b³
  2. (a – b)³ = a³ + 3ab(a – b) – b³
  3. (a + b)³ – (a³ + b³) = 3ab(a + b)
  4. (a – b)³ – (a³ – b³) = 3ab(a – b)
  5. a³ + b³ = (a + b)³ – 3ab(a + b)
  6. a³ – b³ = (a – b)³ + 3ab(a – b)

See More:

Expansion of (a ± b)^3 Examples

Example 1.
Solve the equation (x – 2y)³ using (a – b)³ formula
Solution:
Given that
(x – 2y)³
We know that
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here
a = 1, b= -2y
Substitute a, b in the above equation then we get
x³ – 3(x)² (-2y) + 3(x)(-2y)² – (-2y)³
x³ + 6x²y – 12xy² + 8y³
Therefore the solution of the expression (x – 2y)³ is x³ + 6x²y – 12xy² + 8y³

Example 2.
Solve the equation (2x + 4y)³ using (a – b)³ formula
Solution:
Given that
(2x + 4y)³
We know that
(a – b)³ = a³ – 3a²b + 3ab² – b³
Here
a = 2x, b= 4y
Substitute a, b in the above equation then we get
2x³ – 3(2x)² (4y) + 3(2x)(4y)² – (4y)³
2x³ – 24x²y + 24xy² – 64y³
Therefore the solution of the expression (2x + 4y)³ is 2x³ – 24x²y + 24xy² – 64y³

Example 3.
Solve the equation (2x + y)³ using the formula (a + b)³
Solution:
Given that
(2x + y)³
We know that
(a + b)³ = a³ + 3ab(a + b) + b²
Here a = 2x, b = y
Substitute a, b in the above equation then we get
(2x)³ + 3(2x)(y)(2x + y) + y³
2x³ + 6xy(2x + y) + y³
2x³ + 12x²y + 6xy² + y³
Therefore the solution of the expression (2x + y)³ is 2x³ + 12x²y + 6xy² + y³

Example 4.
Solve the equation (3x + 2y)³ using the formula (a + b)³
Solution:
Given that
(3x + 2y)³
We know that
(a + b)³ = a³ + 3a²b + 3ab² + b³
Here
a= 3x, b= 2y
Substitute a, b in the above equation then we get
(3x)³ + 3(3x)²(2y) + 3(3x)(2y)² + (2y)³
27x³ + 3(9x²)(2y) + 3(3x)(4y²) + 8y³
27x³ + 54x²y + 36xy² + 8y³
Therefore the solution of the expression (2x + y)³ is 27x³ + 54x²y + 36xy² + 8y³

Example 5.
Solve the equation (4x + 2y)³ using the formula (a + b)³
Solution:
Given that
(4x + 2y)³
We know that
(a + b)³ = a³ + 3ab(a + b) + b²
Here a = 4x, b = 2y
Substitute a, b in the above equation then we get
(4x)³ + 3(4x)(2y)(4x + 2y) + 2y³
4x³ + 24xy(4x + 2y) + 2y³
4x³ + 96x²y + 48xy² + 2y³
Therefore the solution of the expression (4x + 2y)³ is 4x³ + 96x²y + 48xy² + 2y³

Worksheet on Application Problems on Expansion of Powers of Binomials and Trinomials

Worksheet on Application Problems on Expansion of Powers of Binomials and Trinomials | Applications of Binomial and Trinomial Expansion of Powers Worksheet

Students who would like to practice the problems on Expansion of Powers of Binomials and Trinomials can get them here. We have provided the Expansion of Powers of Binomials and Trinomials in the below section. There are many benefits for the students by referring to our  Problems on Expansion of Powers of Binomials and Trinomials Worksheet pdf. Relate the concept to real-world problems to understand the topic in depth. This will help you to improve your math skills and also score top in the class.

Read Similar:

Application Problems on Expansion of Powers of Binomials and Trinomials Worksheet

Example 1.
If the sum of two numbers is 5 and the difference between the two numbers is 3. Find the squares of difference of two numbers.
Solution:
Given,
The sum of two numbers is 5 and
The difference between the two numbers is 3
a + b = 5
a – b = 3
By using the formula (a + b)(a – b) = a² – b²
a² – b² = (5)(3)
a² – b² = 15

Example 2.
Evaluate the following using (x + y)(x – y) = x² – y²
i. 13 × 7
ii. 12 × 8
iii. 23 × 17
Solution:
i. 13 × 7
By using the formula (a + b)(a – b) = a² – b²
(10 + 3) (10 – 3) = 10² – 3²
= 100 – 9
= 91
ii. 12 × 8
By using the formula (a + b)(a – b) = a² – b²
(10 + 2)(10 – 2) = 10² – 2²
= 100 – 4
= 96
iii. 23 × 17
By using the formula (a + b)(a – b) = a² – b²
(20 + 3)(20 – 3) = 20² – 3²
= 400 – 9
= 391

Example 3.
If the sum of the squares of the two numbers is 10 and the sum of the two terms is 5. Find the product of the two constant terms.
Solution:
Given,
The sum of the squares of the two numbers is 10 and
The sum of the two terms is 5.
x + y = 5
x² + y² = 10
This is in the form of (x + y)² = x² + 2xy + y²
(5)² = 10 + 2xy
25 = 10 + 2xy
2xy = 25 – 10
2xy = 15
xy = 15/2
xy = 7.5
Thus the product of the two constant terms is 7.5

Example 4.
Evaluate the following using (x ± y)² = x² ± 2xy + y²
i. 37²
ii. (4.03)²
iii. 16²
Solution:
i. 37²
We can write it as (40 – 3)²
This is in the form of (x – y)² = x² – 2xy + y²
(40 – 3)² = 40² – 2(40)(3) + 3²
= 1600 – 240 + 9
= 1369
ii. (4.03)²
We can write it as (4 + 0.03)²
This is in the form of (x + y)² = x² + 2xy + y²
(4 + 0.03)² = 4² + 2(4)(0.03) + (0.03)²
= 16 + 0.24 + 0.0009
= 16.2409
iii. 16²
We can write it as (20 – 4)²
This is in the form of (x – y)² = x² – 2xy + y²
(20 – 4)² = 20² – 2(20)(4) + 4²
= 400 – 160 + 16
= 256

Example 5.
Evaluate the number 6.12 × 5.88
Solution:
Given,
6.12 × 5.88
We can write it as, (6 + 0.12) × (6 – 0.12)
By using the formula (a + b)(a – b) = a² – b²
(6 + 0.12) (6 – 0.12) = 6² – 0.12²
= 36 – 0.0144
= 35.98

Example 6.
If the sum of the three numbers a, b, c is 4 and the sum of their squares is 20 then find the sum of the product of the three numbers taking two at a time.
Solution:
Given that,
The sum of the three numbers a, b, c is 4
a + b + c = 4
The sum of their squares is 20
a² + b² + c² = 20
We need to find the value of ab + bc + ca
We know that
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b + c)² – a² + b² + c² = 2(ab + bc + ca)
4² – 20 = 2( ab + bc + ca)
16 – 20 = 2( ab + bc + ca)
-4 = 2(ab + bc + ca)
ab + bc + ca = -4/2
Thus ab + bc + ca = -2

Example 7.
If the sum of two numbers is 7 and the sum of their cubes is 28, find the sum of their squares.
Solution:
Given,
The sum of two numbers is 7 and
The sum of their cubes is 28
a + b = 7
a³ + b³ = 28
a³ + b³ = (a + b)³ – 3ab(a + b)
28 = (7)³ – 3ab(7)
28 = 343 – 21ab
21ab = 343 – 28
21ab = 315
ab = 15
Now a² + b² = (a + b)² – 2ab
a² + b² = (6)² – 2(15)
a² + b² = 36 – 30
a² + b² = 6

Example 8.
Use (x ± y)² = x² ± 2xy + y² to evaluate (2.06)²
Solution:
Given
(2.06)²
We can write the given number as (2 + 0.06)²
This is in the form of (x + y)² = x² + 2xy + y²
(2 + 0.06)² = 2² + 2(2)(0.06) + (0.06)²
= 4 + 0.24 + 0.0036
= 4.2436
Thus the value of (2.06)² is 4.2436

Example 9.
Find the value of (3.98)²
Solution:
Given,
(3.98)²
We can write the given number as (4 – 0.2 )²
This is in the form of (x – y)² = x² – 2xy + y²
(4 – 0.2 )² = 4² – 2(4)(0.2) + (0.2)²
= 16 – 0.16 + 0.04
= 15.88
Thus the value of (3.98)² is 15.88

Example 10.
If the sum of two numbers x and y is 15 and the sum of their squares is 35 find the product of the numbers.
Solution:
Given,
The sum of two numbers x and y is 15 and
The sum of their squares is 35
x + y = 15
x² + y² = 35
This is in the form of (x + y)² = x² + 2xy + y²
(15)² = 35 + 2xy
225 = 35 + 2xy
225 – 35 = 2xy
190 = 2xy
xy = 190/2
xy = 95
Therefore the product of the numbers is 95.

Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries

Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries | Expansion of Trinomial with Power 2 and its Corollaries Worksheet

Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries are available for free here. Practice the questions on Expansion of (a ± b ± c)² related to the Expansion of Powers of Binomials and Trinomials concept and improve your knowledge. The square of the constant terms cannot be a negative number. Students who are lagging in expanding the trinomial expressions can follow our page.

Formula:
i. (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
ii. (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)

See More:

Worksheet on Expanding of (a ± b ± c)^2 and its Corollaries

Learn the formulas and their corollaries and know how to solve the problems.

Example 1.
Expand the following trinomial squares.
i. (3x + 2y + z)²
ii. (x + y + 2z)²
iii. (a – 2b – 3c)²
Solution:
i. (3x + 2y + z)²
We have to substitute the values of a, b, c in the formula (a + b + c)²
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(3x + 2y + z)² = (3x)² + (2y)² + z² + 2((3x)(2y) + (2y)z + z(3x))
(3x + 2y + z)² = 9x² + 4y² + z² + 2(6xy + 2yz + 3xz)
(3x + 2y + z)² = 9x² + 4y² + z² + 12xy + 4yz + 6xz
Thus the expansion of (3x + 2y + z)² is 9x² + 4y² + z² + 12xy + 4yz + 6xz
ii. (x + y + 2z)²
We have to substitute the values of a, b, c in the formula (a + b + c)²
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(x + y + 2z)² = (x)² + (y)² + (2z)² + 2((x)(y) + y(2z) + (2z)(x))
(x + y + 2z)² = x² + y² + 4z² + 2xy + 2yz + 4zx
Thus the expansion of (x + y + 2z)² is x² + y² + 4z² + 2xy + 2yz + 4zx
iii. (a – 2b – 3c)²
Given the trinomial expression (a – 2b – 3c)²
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(a – 2b – 3c)² = a² + (2b)² + (3c)² – 2(a(2b) – 2b(3c) + 3ca)
(a – 2b – 3c)² = a² + 4b² + 9c² – 4ab – 12bc + 6ca
Thus the expansion of (a – 2b – 3c)² is a² + 4b² + 9c² – 4ab – 12bc + 6ca

Example 2.
If a + b + c = 5 and ab + bc + ca = 10, find a² + b² + c².
Solution:
Given,
a + b + c = 5
ab + bc + ca = 10
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
Substitute the given values in the formula.
(5)² = a² + b² + c² + 2(10)
25 = a² + b² + c² + 20
25 – 20 = a² + b² + c²
a² + b² + c² = 5

Example 3.
Expand (a + 4b + 5c)² by using the (a + b + c)² formula.
Solution:
Given (a + 4b + 5c)²
We have to substitute the values of a, b, c in the formula (a + b + c)²
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + 4b + 5c)² = a² + (4b)² + (5c)² + 2(a(4b) + (4b)(5c) + (5c)a)
(a + 4b + 5c)² = a² + 16b² + 25c² + 2(4ab + 20bc + 5ca)
(a + 4b + 5c)² = a² + 16b² + 25c² + 8ab + 40bc + 10ca
Thus the expansion of (a + 4b + 5c)² is a² + 16b² + 25c² + 8ab + 40bc + 10ca

Example 4.
Find the value of (x + y + z)² if the values of x, y, z are 4, 5, 6.
Solution:
The values of x, y, z are 4, 5, 6.
We have to substitute the values of x, y, z in the formula.
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(4 + 5 + 6)² = 4² + 5² + 6² + 2(4(5) + (5)(6) + (6)(4))
(4 + 5 + 6)² = 16 + 25 + 36 + 2(20 + 30 + 24)
(4 + 5 + 6)² = 16 + 25 + 36 + 2(74)
(4 + 5 + 6)² = 16 + 25 + 36 + 2(74)
(4 + 5 + 6)² = 225
Therefore the value of the trinomial expression (4 + 5 + 6)² = 225

Example 5.
Find the value of the trinomial expression (a + b + c)² if the values of a, b, c are 7, 4, 3.
Solution:
Given the values of a, b, c are 7, 4, 3.
We have to substitute the values of a, b, c in the formula (a + b + c)²
We know that,
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(7 + 4 + 3)² = 7² + 4² + 3² + 2(7(4) + 4(3) + 3(7))
(7 + 4 + 3)² = 49 + 16 + 9 + 2(28 + 12 + 21)
(7 + 4 + 3)² = 49 + 16 + 9 + 2(61)
(7 + 4 + 3)² = 196
Therefore the value of the trinomial expression (a + b + c)² is 196.

Example 6.
Expand the expression (a – b – c)² if a = 2, b = 4, c = 6
Solution:
a = 2, b = 4, c = 6
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(2 – 4 – 6)² = 2² + 4² + 6² – 2(2(4) – 4(6) + 6(2))
(2 – 4 – 6)² = 4 + 16 + 36 – 2(8 – 24 + 12)
(2 – 4 – 6)² = 4 + 16 + 36 – 2(-4)
(2 – 4 – 6)² = 4 + 16 + 36 + 8
(2 – 4 – 6)² = 4 + 16 + 36 + 8 = 64
Thus the value of expression (a – b – c)² is 64

Example 7.
If a + b + c = 12 and ab + bc + ca = 22 then find a² + b² + c²
Solution:
Given,
a + b + c = 12 and ab + bc + ca = 22
We know that,
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
Substitute the given values in the formula.
(12)² = a² + b² + c² + 2(22)
144 = a² + b² + c² + 44
144 – 44 = a² + b² + c²
Thus the value of a² + b² + c² is 144.

Example 8.
If a = 5, b = 6 and c = 7 find the value of the following trinomial expressions
i. (a – b – c)²
ii. (a + b + c)²
Solution:
i. (a – b – c)²
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(5 – 6 – 7)² = 5² + 6² + 7² – 2(5(6) – 6(7) + 7(5))
(5 – 6 – 7)² = 25 + 36 + 49 – 2(30 – 42 + 35)
(5 – 6 – 7)² = 64
Thus the value of (5 – 6 – 7)² is 64
ii. (a + b + c)²
We know that,
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
We have to substitute the values of a, b, c in the formula (a + b + c)²
(5 + 6 + 7)² = 5² + 6² + 7² + 2(5(6) + 6(7) + 7(5))
(5 + 6 + 7)² = 25 + 36 + 49 + 2(30 + 42 + 35)
(5 + 6 + 7)² = 324
Thus the value of (5 + 6 + 7)² is 324

Example 9.
Expand the squares of the trinomials 1 – x – x²
Solution:
Given,
1 – x – x²
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(1 – x – x²)² = 1² + x² + (x²)² – 2(1(x) – x(x²) + x²(1))
(1 – x – x²)² = 1 + x² + (x²)² – 2(x – x³ + x²)
Thus the expansion of the squares of the trinomials 1 – x – x² is 1 + x² + (x²)² – 2(x – x³ + x²)

Example 10.
Expand the squares of the trinomials x + y – z
Solution:
Given,
x + y – z
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(x – (-y) – z)² = x² + (-y)² + (z)² – 2(x(-y) – (-y)z + z(x))
(x – (-y) – z)² = x² + y² + z² – 2(-xy + yz + zx)
(x – (-y) – z)² = x² + y² + z² + 2xy + 2yz + 2zx
(x + y – z)² = x² + y² + z² + 2xy + 2yz + 2zx
Thus the expansion of the squares of the trinomials (x + y – z)² is x² + y² + z² + 2xy + 2yz + 2zx

Worksheet on Expansion of (a ± b)^2 and its Corollaries

Worksheet on Expansion of (a ± b)^2 and its Corollaries | Expanding (a ± b)^2 and its Corollaries Worksheet with Answers

All the students who are unable to understand the concept of Expansion of Powers of Binomials and Trinomials can refer to our page and learn the topics quickly and easily. Enhance your math skills by solving the problems from Worksheet on Expansion of (a ± b)^2 and its Corollaries. Practice the questions from here and secure the highest marks in the exams. Step-by-step explanations for all the questions are provided in the simple techniques in this Worksheet on Expansion of (a ± b)² and its Corollaries.

Also, See:

Expansion of (a ± b)^2 and its Corollaries Worksheet PDF

Check out the questions given below and try to solve the problems quickly.

Example 1.
Simplify (4m + 6n)² + (4m – 6n)²

Solution:

Given that
(4m + 6n)² + (4m – 6n)²
We know that
The formula of (a + b)² + (a – b)² = 2(a² + b²)
2{(4m)² + (6n)²}
2(16m² + 36n²)
32m² + 72n²
Thus (4m + 6n)² + (4m – 6n)² is 32m² + 72n²


Example 2.
Expand (4a + 1/4a)² by using (a + b)² formula.

Solution:

Given that
(4a + 1/4a)²
We know that
The formula of (a + b)² + (a – b)² = 2(a² + b²)
(4a)² + 4 × 4p × 1/4p + (1/4p)²
16a² + 4 + (1/16p²)
Thus the expansion of (4a + 1/4a)² is 16a² + 4 + (1/16p²)


Example 3.
Expand (5a + 2b)² by using the (a + b)² formula.

Solution:

Given that
(5a + 2b)²
We know that
The formula of (a + b)² + (a – b)² = 2(a² + b²)
(5a)² + 2 × 5a × 2b + (2b)²
25a² + 20ab + 4b²
Thus the expansion of (5a + 2b)² is 25a² + 20ab + 4b²


Example 4.
Simply the equation (a + 1/a)² + (a + 1/a)².

Solution:

We know that
The formula of (a + b)² + (a – b)² = 2(a² + b²)
2(a² – 1/a² + a² + 1/a²)
2(a² – 1/a² + a² + 1/a²)
2(2a²)
4a²
Thus (a + 1/a)² + (a + 1/a)² is 4a²


Example 5.
If a + b = 4 and ab = 2. Find a² + b²

Solution:

We know that
The formula of a² + b² = (a + b)² – 2ab
a² + b² = (4)² – 2(2)
16 – 4
12
Thus the value of the expression a² + b² is 12.


Example 6.
If x + y = 8 and x – y = 4 evaluate xy

Solution:

Given,
x + y = 8 and x – y = 4
We know that
xy = (x + y)(x – y)
xy = 8 × 4
xy = 32
Therefore the value of xy is 32.


Example 7.
If a + 1/a = 2 find the value of a⁴ + 1/a⁴.

Solution:

Given,
a + 1/a = 2
We notice that
(a + 1/a)² = 2² = 4
But
(a + 1/a)² = a² + (1/a)² + 2 × a(1/a)
= a² + 1/a² + 2 = 4
a² + 1/a² = 4 – 2 = 2
Again
(a² + 1/a²)² = 2² = 4
Know
(a² + 1/a²)² = a⁴ + 2(a²)(1/a)² + 1/a⁴
a⁴ + 2 + 1/a⁴
a⁴ + 1/a⁴ + 2 = 4
a⁴ + 1/a⁴ = 2
Thus a⁴ + 1/a⁴ = 2


Example 8.
Expand the equation (2x – 3y)² by using the (a – b)² formula.

Solution:

Given that
(2x – 3y)²
We know that
The formula of (a – b)² = a² + b² + 2ab
(2x)² + (3y)² – 2(2x)(3y)
4x² + 9y² – 12xy
Thus (2x – 3y)² is 4x² + 9y² – 12xy


Example 9.
Expand the squares of ½ + 4/2n by using the (a + b)² formula.

Solution:

Given that
½ + 4/2 n
We know that
The formula of (a + b)² = a² + b² + 2ab
(½)² + (4/2 n)² + 2 × ½ × 4/2 n
¼ + 16/4 n² + 2n
¼ + 4n² + 2n
¼ + 2n + 4n²
Thus the squares of ½ + 4/2n is ¼ + 2n + 4n²


Example 10.
Expand the squares of the 2x + y by using the (a + b)² formula.

Solution:

Given that,
2x + y
Square of the 2x + y = (2x + y)²
We know that
The formula of (a + b)² = a² + b² + 2ab
(2x)² + y² + 2xy
4x² + 2xy + y²
Thus squares of the 2x + y is 4x² + 2xy + y²


Application Problems on Expansion of Powers of Binomials and Trinomials

Application Problems on Expansion of Powers of Binomials and Trinomials | Binomial and Trinomial Expansion Question and Answers

Application Problems on Expansion of Powers of Binomials and Trinomials are available here. So, the students who are in search of problems related to the expansion of powers of binomials and trinomials can use this page and practice the sums. In this article, you can find different types of problems on the expansion of powers of binomials and trinomials with step-by-step explanations. Practice the given questions and test your knowledge on this topic.

Do Refer:

Application Problems on Expansion of Powers of Binomials and Trinomials

Students of 9th grade can make use of the below examples and improve their math skills.

Example 1.
Use (x ± y)² = x² ± 2xy + y² to evaluate (3.07)²
Solution:
(3.07)²
We can evaluate (3.07)² by using (x + y)² = x² + 2xy + y²
(3 + 0.07)²
3² + 2 × (3)(0.07) + (0.07)²
9 + 2 × 0.21 + 0.0049
9 + 0.42 + 0.0049
9.4249

Example 2.
Use (x ± y)² = x² ± 2xy + y² to evaluate (8.92)².
Solution:
(8.92)²
We can evaluate (8.92)² by using (x – y)² = x² – 2xy + y²
(9 – 0.08)²
9² – 2(9)(0.08) + (0.08)²
81 – 1.44 + 0.0064
81 – 1.4464
79.553

Example 3.
Evaluate 24 × 16 using (x + y)(x – y) = x² – y²
Solution:
(20 + 4)(20 – 4)
20² – 4²
400 – 16
384

Example 4.
Evaluate 6.98 × 7.02
Solution:
We can evaluate 6.98 × 7.02 by using the formula (x + y)(x – y) = x² – y²
(7 – 0.02)(7 + 0.02)
7² – 0.02²
49 – 0.0004
48.99

Example 5.
If the sum of two numbers x and y is 20 and the sum of their squares is 56 find the product of the numbers.
Solution:
Sum of two numbers x and y is 20
Sum of the two numbers x and y squares is 56
x² + y² = 56
We know that
2ab = (a + b)² – (a² + b²)
2xy = 20² – 56
200 – 56
2xy = 144
Therefore xy = 1/2 × 2xy
1/2 × 144 = 72.

Example 6.
If the sum of three numbers a, b, c is 3 and the sum of their squares is 12 then find the sum of the products of the three numbers taking two at a time.
Solution:
Sum of three numbers a, b, c is 3
a + b + c = 3
Sum of three numbers a, b, c squares is 12
a² + b² + c² = 12
We need to find the value of ab + bc + ca
We know that
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(a + b + c)² – a² + b² + c² = 2(ab + bc + ca)
3² – 12 = 2( ab + bc + ca)
9 – 12 = 2( ab + bc + ca)
-3 = 2(ab + bc + ca)
ab + bc + ca = -2/3

Example 7.
Evaluate (2.34)² + (3.14)²
Solution:
a³ + b³ = (a + b)³ – 3ab(a + b)
(2.34)³ + (3.14)³
(2.34 + 3.14)³ – 3(2.34)(3.14)(2.34+3.14)
164.56 – 120.79
43.77

Example 8.
If the sum of two numbers is 6 and the sum of their cubes is 27, find the sum of their squares
Solution:
Given,
Sum of the numbers is 6
The two numbers are a, b.
a + b = 6
Sum of their cubes is 27
a³ + b³ = 27
a³ + b³ = (a + b)³ – 3ab(a + b)
6³ – 27 = 3ab(a + b)
18ab = 216 – 27 = 189
ab = 189/18 = 23.625
Now a² + b² = (a + b)² – 2ab
6² – 2 × 23. 625
36 – 2 × 23.625
803.26

Example 9.
Use (x ± y)² = x² ± 2xy + y² to evaluate (1.07)²
Solution:
Given that,
(1.07)²
(1 + 0.07)²
1² + 2 × (1)(0.07) + (0.07)²
1 + 2 × 0.07 + 0.0049
1 + 0.14 + 0.0049
1.1449

Example 10.
If the sum of two numbers x and y is 5 and the sum of their squares is 12 find the product of the numbers.
Solution:
Sum of two numbers x and y is 5
Sum of the two numbers x and y squares is 12
x² + y² = 12
We know that
2ab = (a + b)² – (a² + b²)
2xy = 5² – 12
25 – 12
2xy = 13
Therefore xy = 1/2 × 2xy
1/2 × 13 = 13/2.

Expansion of (x ± a)(x ± b) ? – Derivation, Formulas | How to Expand (x ± a)(x ± b)?

According to the binomial theorem, the sum and product of the constant and variable terms will be a quadratic equation. The expansion of (x ± a)(x ± b) is the individual product of the variable x and constant terms. It can be read as the square of x and the sum of the constant terms and product of the constant terms. Read the entire article to derive (x + a)(x + b), (x – a)(x – b), (x + a)(x – b) and (x – a)(x + b). In addition to the derivations, the students can find examples of Expansion of Powers of Binomials and Trinomials.

Expansion of (x ± a)(x ± b)

Let us discuss briefly the expansion of (x ± a)(x ± b) with proves from here.

What is the expansion of (x + a)(x + b)?
The derivation of expanding (x + a)(x + b) is given below,
(x + a)(x + b) = x(x + b) + a(x + b)
(x + a)(x + b) = x² + bx + ax + ab
(x + a)(x + b) = x² + x(a + b) + ab

What is the expansion of (x – a)(x – b)?
The derivation of expanding (x – a)(x – b) is given below,
(x – a)(x – b) = x(x – b) – a(x – b)
(x – a)(x – b) = x² – bx – ax + ab
(x – a)(x – b) = x² – x(a + b) + ab

What is the expansion of (x + a)(x – b)?
Derivation on how to expand (x + a) (x – b) is as follows,
(x + a)(x – b) = x(x – b) + a(x – b)
(x + a)(x – b) = x² – bx + ax – ab
(x + a)(x – b) = x² + x(a – b) – ab

What is the expansion of (x – a)(x + b)?
The derivation of expanding (x – a)(x – b) is given below,
(x – a)(x + b) = x(x + b) – a(x + b)
(x – a)(x + b) = x² + bx – ax – ab
(x – a)(x + b) = x² – x(a – b) – ab

Formulas of (x ± a)(x ± b)

  1. (x – a)(x + b) = x² – x(a – b) – ab
  2. (x + a)(x – b) = x² + x(a – b) – ab
  3. (x – a)(x – b) = x² – x(a + b) + ab
  4. (x + a)(x + b) = x² + x(a + b) + ab

See More:

Solved Problems on Expansion of (x ± a)(x ± b)

Example 1.
Find the product of (x + 2) (x + 4) using the standard formula.
Solution:
Given the binomial expression (x + 2) (x + 4)
By using the formula (x + a)(x + b) = x² + x(a + b) + ab we can expand the given expression.
(x + 2) (x + 4) = x² + x(2 + 4) + (2)(4)
(x + 2) (x + 4) = x² + 6x+ 8
Thus the product of (x + 2) (x + 4) is x² + 6x+ 8

Example 2.
Expand (a – 5)(a – 10) using the (x ± a)(x ± b) formula.
Solution:
Given the binomial expression (a – 5)(a – 10)
By using the formula (x – a)(x – b) = x² – x(a + b) + ab
(a – 5)(a – 10) = a² – a(5 + 10) + 5(10)
(a – 5)(a – 10) = a² – 15a + 50
Thus the expansion of (a – 5)(a – 10) is a² – 15a + 50

Example 3.
Find the product of (m + 5)(m + 1) using the standard formula.
Solution:
Given,
(m + 5)(m + 1)
By using the formula (x + a)(x + b) = x² + x(a + b) + ab we can expand the given expression.
(m + 5)(m + 1) = m² + m(5 + 1) + 5
(m + 5)(m + 1) = m² + m(6) + 5
(m + 5)(m + 1) = m² + 6m + 5
Thus the expansion of (m + 5)(m + 1) is m² + 6m + 5

Example 4.
Find the product of (x + 2)(x – 3) using the standard formula.
Solution:
Given the binomial expression (x + 2)(x – 3)
By using the formula (x + a)(x – b) = x² + x(a – b) – ab we can expand the given expression.
(x + 2)(x – 3) = x² + x(2 – 3) – 2(3)
(x + 2)(x – 3) = x² + x(-1) – 2(3)
(x + 2)(x – 3) = x² – x – 6

Example 5.
Find the product of (2m + 1)(2m – 4) using the standard formula.
Solution:
Given the binomial expression (2m + 1)(2m – 4)
By using the formula (x + a)(x – b) = x² + x(a – b) – ab we can expand the given expression.
(2m + 1)(2m – 4) = (2m)² + 2m(1 – 4) – 1(4)
(2m + 1)(2m – 4) = 4m² + 2m – 8m – 4
(2m + 1)(2m – 4) = 4m² – 6m – 4

FAQs on Expansion of (x ± a)(x ± b)

1. What is the expansion of (x + a)(x + b)?
The expansion of (x + a)(x + b) is x² + x(a + b) + ab

2. What is the expansion of (x + a)(x – b)?
The expansion of (x + a)(x – b) is x² + x(a – b) – ab

3. How to expand (x + a)(x + b)?
(x + a)(x + b) = x(x + b) + a(x + b)
(x + a)(x + b) = x² + bx + ax + ab
(x + a)(x + b) = x² + x(a + b) + ab

Expansion of (a ± b ± c)^2

Expansion of (a ± b ± c)^2 | Expansion of Trinomial with Power 2 Derivation, Examples

The Expansion of (a ± b ± c)^2 can be read as the whole squares of a plus or minus b plus or minus c. It is the formula that is used to find the square of sum or difference among three constant terms such as a, b, c. To derive the Expansion of Powers of Binomials and Trinomials we need to use some set of formulas. Let us learn the derivations of (a + b + c)² and (a – b – c)² with some suitable examples from this page.

Also, Read: Expansion of (a ± b)^2

Expansion of (a ± b ± c)^2 | Expansion of Trinomial with Power 2

Expansion with three constant terms like a, b, c is known as the trinomial expressions. Check the below section to know how to derive (a + b + c)² and (a – b – c)² and where it is used.

Derivations of (a + b + c)²: 
The derivation of (a + b + c)² is explained in detail here.
(a + b + c)² = (a + b + c) (a + b + c)
(a + b + c)² = a (a + b + c) + b (a + b + c) + c (a + b + c)
(a + b + c)² = a² + ab + ac + ab + b² + bc + ac + bc + c²
(a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ac
(a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
Thus the formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

Derivation of (a – b – c)²:
The derivation of the trinomial expression (a – b – c)² is explained in detail here.
(a – b – c)² = (a – b – c) (a – b – c)
(a – b – c)² = a (a – b – c) – b (a – b – c) -c (a – b – c)
(a – b – c)² = a² – ab – ac – ab + b² + bc – ac – bc + c²
(a – b – c)² = a² + b² + c² – 2ab + 2bc – 2ac
(a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
Thus the formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)

Expansion of (a ± b ± c)^2 Examples

Let us see some suitable examples on the expansion of (a ± b ± c)² to know how and where to use the formulas of (a + b + c)² and (a – b – c)² in algebra.

Example 1.
Find the value of (a + b + c)² if the values of a, b, c are 3, 4, 2.
Solution:
Given the values of a, b, c are 3, 4, 2.
We have to substitute the values of a, b, c in the formula (a + b + c)²
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
(3 + 4 + 2)² = 3² + 4² + 2² + 2(3(4) + (4)2 + 2(3))
= 9 + 16 + 4 + 2(12 + 8 + 6)
= 29 + 2(26)
= 29 + 52
= 81

Example 2.
Expand (2a + 4b + 3c)² by using the (a + b + c)² formula.
Solution:
Given the trinomial expression (2a + 4b + 3c)²
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
We have to substitute the values of a, b, c in the formula (a + b + c)²
(2a + 4b + 3c)² = (2a)² + (4b)² + (3c)² + 2((2a)(4b) + (4b)3c + 3c(2a))
(2a + 4b + 3c)² = 4a² + 16b² + 9c² + 2(8ab + 12bc + 6ac)
(2a + 4b + 3c)² = 4a² + 16b² + 9c² + 16ab + 24bc + 12ac
Thus the expansion of (2a + 4b + 3c)² is 4a² + 16b² + 9c² + 16ab + 24bc + 12ac

Example 3.
Find the value of (a – b – c)² if the values of a, b, c are 5, 3, 2.
Solution:
Given the values of a, b, c are 5, 3, 2.
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(5 – 3 – 2)² = 5² + 3² + 2² – 2((5)(3) – (3) (2) + (2)(5))
(5 – 3 – 2)² = 25 + 9 + 4 – 2(15 – 6 + 10)
(5 – 3 – 2)² = 38 – 2(19)
(5 – 3 – 2)² = 38 – 38 = 0
Thus the value of (a – b – c)² if the values of a, b, c are 5, 3, 2 is 0.

Example 4.
Expand (3x + 2y + 5z)² by using the (a + b + c)² formula.
Solution:
Given the trinomial expression (3x + 2y + 5z)²
We know that
The formula of (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)
We have to substitute the values of a, b, c in the formula (a + b + c)²
(3x + 2y + 5z)² = (3x)² + (2y)² + (5z)² + 2((3x)(2y) + (2y)(5z) + (5z)(3x))
(3x + 2y + 5z)² = 9x² + 4y² + 25z² + 2(6xy + 10yz + 15zx)
(3x + 2y + 5z)² = 9x² + 4y² + 25z² + 12xy + 20yz + 30zx
Thus the expansion of (3x + 2y + 5z)² is 9x² + 4y² + 25z² + 12xy + 20yz + 30zx

Example 5.
Expand (8x – 4y – 6z)² by using the (a – b – c)² formula.
Solution:
Given the trinomial expression (8x – 4y – 6z)²
We know that
The formula of (a – b – c)² = a² + b² + c² – 2(ab – bc + ca)
We have to substitute the values of a, b, c in the formula (a – b – c)²
(8x – 4y – 6z)² = (8x)² + (4y)² + (6z)² – 2((8x)(4y) – (4y)(6z) + (6z)(8x))
(8x – 4y – 6z)² = 64x² + 16y² + 36z² – 2(32xy) – 24yz + 48zx)
(8x – 4y – 6z)² = 64x² + 16y² + 36z² – 64xy – 48yz + 96zx
Thus the expansion of (8x – 4y – 6z)² is 64x² + 16y² + 36z² – 64xy – 48yz + 96zx

FAQs on Expansion of (a ± b ± c)²

1. What is the Expansion of the (a – b – c)² Formula?

The Expansion of (a – b – c)² formula is a² + b² + c² – 2(ab – bc + ca) or a² + b² + c² – 2ab + 2bc – 2ac

2. What is the Expansion of (a + b + c)² Formula?

The Expansion of (a + b + c)² formula is a² + b² + c² + 2(ab + bc + ca) or a² + b² + c² + 2ab + 2bc + 2ac

3. What type of expression is Expansion of (a ± b ± c)^2?

The Expansion of (a ± b ± c)^2 is a trinomial expression because it consists of three constant terms.