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Home >> Standard Identities & their applications >> (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca >>

## (A + B + C) Whole Square

 (a + b)2 = a2 + b2 + 2ab (a - b)2 = a2 + b2 - 2ab a2 - b2 = (a + b) (a – b) (x + a) (x + b) = x2 + x(a + b) + ab (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca (a + b)3 = a3 + b3 + 3ab(a + b) (a - b)3 = a3 - b3 - 3ab(a - b) a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Before you understand (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca, you are advised to read:

How to multiply Variables ?
How to multiply Constant and Variable ?
Multiplication of Polynomials ?
What is Exponential Form and Laws of Exponents ?

How identity of (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca
is obtained

Taking LHS of the identity:
(a + b +c)2
This can also be written as:
= (a + b + c) (a + b + c)

Multiply as we do multiplication of trinomials and we get:
= a(a + b + c) + b(a + b + c) + c(a + b + c)
= a2 + ab + ac + ab + b2 + bc + ac + bc + c2

Rearrange the terms and we get:
= a2 + b2 + c2 + ab + ab + bc + bc + ac + ac

Add like terms and we get:
= a2 + b2+ c2 + 2ab + 2bc + 2ca

Hence, in this way we obtain the identity i.e. (a + b + c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca

Following are the few examples of this identity

Example 1: Solve (4p + 5q + 3r)2
Solution: This proceeds as:
Given polynomial (4p + 5q + 3r)2 represents identity first i.e. (a + b+ c)2
Where a = 4p, b = 5q and c = 3r

Now apply values of a, b and c on the identity i.e. (a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca and we get:
(4p + 5q + 3r)2 = (4p)2 + (5q)2 + (3r)2 + 2(4p)(5q) + 2(5q)(3r) + 2(3r)(4p)

Expand the exponential forms and we get:
= 16p2 + 25q2 + 9r2 + 2(4p)(5q) + 2(5q)(3r) + 2(3r)(4p)

Solve brackets and we get:
= 16p2 + 25q2 + 9r2 + 40pq + 30qr + 24rp

Hence, (4p + 5q + 3r)2 = 16p2 + 25q2 + 9r2 + 40pq + 30qr + 24rp

Example 2: Solve (8x + 4y + 7z)2
Solution: This proceeds as:
Given polynomial (8x + 4y+ 7z)2 represents identity first i.e. (a + b + c)2
Where a = 8x, b = 4y and c = 7z

Now apply values of a, b and c on the identity i.e. (a + b +c)2 = a2 + b2 + c2 + 2ab + 2bc + 2ca and we get:
(8x + 4y+ 7z)2 = (8x)2 + (4y)2 + (7z)2 + 2(8x)(4y) + 2(4y)(7z) + 2(7z)(8x)

Expand the exponential forms and we get:
= 64x2 + 16y2 + 49z2 + 2(8x)(4y) + 2(4y)(7z) + 2(7z)(8x)

Solve multiplication process and we get:
= 64x2 + 16y2 + 49z2 + 64xy + 56yz + 112zx

Hence, (8x + 4y + 7z)2 = = 64x2 + 16y2 + 49z2 + 64xy + 56yz + 112zx