Question

Prove the following identities:
$\sum a^2(b+c)-\sum a^3-2abc=(b+c-a)(c+a-b)(a+b-c)$.

Answer 1

L.H.S

$=\sum a^2(b+c)-\sum a^3-2abc$

$=a^2(b+c)+b^2(a+c)+c^2(a+b)-a^3-b^3-c^3-2abc$

$=a^{2}b+a^{2}c+b^{2}a+b^{2}c+c^{2}a+c^{2}b-a^3-b^3-c^3-2abc$

$=(-a^3+a{b}^2+a{c}^2-2abc)+(-b^3+b{c}^2+a^{2}b)+(-c^3+c{a}^2+c{b}^2)$

Here to make same commom factor we add and subtract the term

$=-a(a^2-b^2-c^2+2bc)+(b(a^2-b^2+c^2)-2b{c}^2+2b^{2}c)+(2b{c}^2-2b^{2}c+c(a^2-c^2+b^2))$

$=-a(a^2-b^2-c^2+2bc)+b(a^2-b^2-c^2+2bc)+c((a^2-b^2-c^2+2bc)$

$=(a^2-b^2-c^2+2bc)(b+c-a)$

To make common factor as $a+c-b$ we can add or subtract as

$=(a^2+ac-ab-ac+ab-b^2+bc+bc-c^2)(b+c-a)$

$=(a^2+ac-ab)+(ab-b^2+bc)+(bc-ac-c^2)$

$=(\left(a(a+c-b)\right)+\left(b(a-b+c)\right)+(-c(a+c-b)))(b+c-a)$

$=(a+b-c)(c+a-b)(b+c-a)$

R.H.S