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Home >> Standard Identities & their applications >> a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) >>

(A Cube + B Cube + C Cube) - 3abc

(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 a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca), 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 this identity of a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) is obtained:

Taking RHS of the identity:
(a + b + c)(a2 + b2 + c2 - ab - bc - ca )

Multiply each term of first polynomial with every term of second polynomial, as shown below:
= a(a2 + b2 + c2 - ab - bc - ca ) + b(a2 + b2 + c2 - ab - bc - ca ) + c(a2 + b2 + c2 - ab - bc - ca )

= { (a X a2) + (a X b2) + (a X c2) - (a X ab) - (a X bc) - (a X ca) } + {(b X a2) + (b X b2) + (b X c2) - (b X ab) - (b X bc) - (b X ca)} + {(c X a2) + (c X b2) + (c X c2) - (c X ab) - (c X bc) - (c X ca)}

Solve multiplication in curly braces and we get:
= a3 + ab2 + ac2 a2b - abc - a2c + a2b + b3 + bc2 - ab2 b2c - abc + a2c + b2c + c3 - abc bc2 - ac2

Rearrange the terms and we get:
= a3 + b3 + c3 + a2b a2b + ac2- ac2 + ab2 - ab2 + bc2 bc2 + a2c - a2c + b2c b2c - abc - abc - abc

Above highlighted like terms will be subtracted and we get:
= a3 + b3 + c3 - abc - abc - abc

Join like terms i.e (-abc) and we get:
= a3 + b3 + c3 3abc

Hence, in this way we obtain the identity i.e. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)

Let's try the following examples of this identity



Example: Solve 8a3 + 27b3 + 125c3 30abc
Solution: This proceeds as:
Given polynomial (8a3 + 27b3 + 125c3 30abc) can be written as:
(2a)3 + (3b)3 + (5c)3 (2a)(3b)(5c)
And this represents identity:
a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca)
Where a = 2a, b = 3b and c = 5c

Now apply values of a, b and c on the L.H.S of identity i.e. a3 + b3 + c3 - 3abc = (a + b + c)(a2 + b2 + c2 - ab - bc - ca) and we get:
= (2a + 3b + 5c) { (2a)2 +(3b)2 + (5c)2 (2a)(3b) (3b)(5c) (5c)(2a) }

Expand the exponential forms and we get:
= (2a + 3b + 5c) {4a2 +9b2 + 25c2 (2a)(3b) (3b)(5c) (5c)(2a) }

Solve brackets and we get:
= (2a + 3b + 5c) {4a2 +9b2 + 25c2 6ab 15bc 10ca}

Hence, 8a3 + 27b3 + 125c3 30abc = (2a + 3b + 5c) (4a2 + 9b2 + 25c2 6ab 15bc 10ca)

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