 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.

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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