Question:
Prove that (a +b +c)3 -a3 -b3 – 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 $\left.a^{3}-b^{3}=(a-b)\left(a^{2}+b^{2}+a b\right)\right]$
$=(b+c)\left[a^{2}+b^{2}+c^{2}+2 a b+2 b c+2 c a+a^{2}+a^{2}+a b+a c\right]$
$-(b+c)\left(b^{2}+c^{2}-b c\right)$
$=(b+c)\left[b^{2}+c^{2}+3 a^{2}+3 a b+3 a c-b^{2}-c^{2}+3 b c\right]$
$=(b+c)\left[3\left(a^{2}+a b+a c+b c\right)\right]$
$=3(b+c)[a(a+b)+c(a+b)]$
$=3(b+c)[(a+c)(a+b)]$
$=3(a+b)(b+c)(c+a)=\mathrm{RHS}$