Prove the following

Question:

Prove that (a +b +c)3 -a3 -b– c3 =3(a +b)(b +c)(c +a).

Solution:

To prove, $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$

$\mathrm{LHS}=\left[(a+b+c)^{3}-a^{3}\right]-\left(b^{3}+c^{3}\right)$

$=(a+b+c-a)\left[(a+b+c)^{2}+a^{2}+a(a+b+c)\right]$

$-\left[(b+c)\left(b^{2}+c^{2}-b c\right)\right]$

[using identity, $a^{3}+b^{3}=(a+b)\left(a^{2}+b^{2}-a b\right)$ and $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$

$=(b+c)\left[a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a+a^{2}+a^{2}+a b+a c\right]$

$-(b+c)\left(b^{2}+c^{2}-b c\right)$

$=(b+c)\left[b^{2}+c^{2}+3 a^{2}+3 a b+3 a c-b^{2}-c^{2}+3 b c\right]$

$=(b+c)\left[3\left(a^{2}+a b+a c+b c\right)\right]$

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

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

$=3(a+b)(b+c)(c+a)=\mathrm{RHS}$

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