Question

Prove the following identities:
$\frac{a^3(b+c)}{(a-b)(a-c)}+\frac{b^3(c+a)}{(b-c)(b-a)}+\frac{a^3(a+b)}{(c-a)(c-b)}=bc+ca+ab$.

Answer 1

L.H.S

$=\frac{a^3(b+c)}{(a-b)(a-c)}+\frac{b^3(c+a)}{(b-c)(b-a)}+\frac{c^3(a+b)}{(c-a)(c-b)}$

$=\frac{a^3(b+c)(b-c)}{(a-b)(a-c)(b-c)}+\frac{b^3(c+a)(c-a)}{(b-c)(b-a)(c-a)}+\frac{c^3(a+b)(a-b)}{(c-a)(c-b)(a-b)}$

$\frac{-(a^3{b}^2-a^3{c}^2+b^3{c}^2-b^3{a}^2+c^3{a}^2-c^3{b}^2)}{(a-b)(b-c)(c-a)}$

$\frac{-(a^3{b}^2-b^3{a}^2+b^3{c}^2-a^3{c}^2+c^3{a}^2-c^3{b}^2)}{(a-b)(b-c)(c-a)}$

$\frac{-(a^2{b}^2(a-b)-c^3(a^3-b^3)-c^3(a^2-b^2))}{(a-b)(b-c)(c-a)}$

$\frac{(a-b)(c^2(a^2+b^2+ab)-a^2{b}^2-c^3(a+b))}{(a-b)(b-c)(c-a)}$

$=(a-b)\frac{a^2{c}^2+c^2{b}^2+ab{c}^2-a^2{b}^2+a{c}^3-b{c}^3}{(a-b)(b-c)(c-a)}$

$=(a-b)\frac{a^2{c}^2-a^2{b}^2+b^2{c}^2-b{c}^3+ab{c}^2-a{c}^3}{(a-b)(b-c)(c-a)}$

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

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

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

$=(a-b)(b-c)(c-a)\frac{bc+ac+ab}{(a-b)(b-c)(c-a)}$

$=bc+ca+ab$

R.H.S