Question

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

Answer 1

When $x+y+z=0$ then ,

$x^3+y^3+z^3=3xyz$

Let $x=a+b-2c$

$y=b+c-2a$

$z=c+a-2b$

$x+y+z=(a+b-2c)+(b+c-2a)+(c+a-2b)$

$x+y+z=0$

therefore ,

$x^3+y^3+z^3=3xyz$

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

L.H.S

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

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

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

$=$ R.H.S