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)

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## 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