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