Question

prove that (a+b+c)3 - (a + b3 + c) = 3(a+b) (b+c) (c+a)​

Answer 1

Step-by-step explanation:

Solution to (a + b + c) ^ 3 - a ^ 3 - b ^ 3 - c ^ 3 = 3(a + b)

(b + c)(c + a)

To prove, (a + b + c) ^ 3 - a ^ 3 - b ^ 3 - c ^ 3 = 3(a + b) * (b + c) * (c + a)

LHS = [(a + b + c) ^ 3 - a ^ 3] - (b ^ 3 + c ^ 3)

=(a+b+c-a)[(a+b+c)^ 2 +a^ 2 +a(a+b+c)]

- [(b + c)(b ^ 2 + c ^ 2 - bc)]

[using identity, a ^ 3 + b ^ 3 = (a + b)(a ^ 2 + b ^ 2 - ab) * a * n * d * epsilon a^ 3 -b^ 3 =(a-b)(a^ 2 +b^ 2 +ab)]

=(b+c)[a^ 2 +b^ 2 +c^ 2 +2ab+2bc+2ca+a^ 2 +ab+ac]

- (b + c) * (b ^ 2 + c ^ 2 - bc)

=(b+c)[b^ 2 +c^ 2 +3a^ 2 +3ab+3ac-b^ 2 -c^ 2 +3bc]

=(b+c)[3(a^ 2 +ab+ac+bc)]

=3(b+c)[a(a+b)+c(a+b)]

=3(b+c)[(a+c)(a+b)]

=3(a+b)(b+c)(c+a)=RHS

Hence proved.