Question

Prove the following identity:
$\frac{a{(b-c)}^2}{(c-a)(a-b)}+\frac{b{(c-a)}^2}{(a-b)(b-c)}+\frac{c{(a-b)}^2}{(b-c)(c-a)}=a+b+c$.

Answer 1

Take L.C.M of the given expression, we get
$\frac{a(b-c{)}^2(b-c)+b(c-a{)}^2(c-a)+c(a-b{)}^2(a-b)}{(c-a)(a-b)(b-c)}$
$=\frac{a(b-c{)}^3+b(c-a{)}^3+c(a-b{)}^3}{(c-a)(a-b)(b-c)}$
$=\frac{a(b-c{)}^3-b(b-c+a-b{)}^3+c(a-b{)}^3}{(c-a)(a-b)(b-c)}$
$=\frac{a(b-c{)}^3-b(b-c{)}^3-b(a-b{)}^3-3b(b-c\left)\right(a-b\left)\right(a-c)+c(a-b{)}^3}{(c-a)(a-b)(b-c)}$
$=\frac{(b-c{)}^3(a-b)+(a-b{)}^3(c-b)+3b(b-c)(c-a)(a-b))}{(b-c)(c-a)(a-b)}$
$=\frac{(b-c)(a-b)\left(\right(b-c{)}^2-(a-b{)}^2)}{(b-c)(c-a)(a-b)}+\frac{3b(b-c)(c-a)(a-b)}{(b-c)(c-a)(a-b)}$
$=\frac{(b-c)(a-b)\left(\right(b-c+a-b\left)\right(b-c-a+b\left)\right)}{(b-c)(c-a)(a-b)}+3b$
$=\frac{(b-c)(a-b)(a-c)(2b-a-c)}{(b-c)(a-b)(c-a)}+3b$
$=-(2b-a-c)+3b$
$=a+b+c$