Question

Prove the identity ${\left\{\right(y-z{)}^2+(z-x{)}^2+(x-y{)}^2\}}^3-54(y-z{)}^2(z-x{)}^2(x-y{)}^2=2(y+z-2x{)}^2(z+x-2y{)}^2(x+y-2z{)}^2$.

Answer 1

Let $y-z=a,z-x=b,x-y=c,$ so that $a+b+c=0$

Then, we have to prove that, $(a^2+b^2+c^2{)}^3-54a^2{b}^2{c}^2=2(b-c{)}^2(c-a{)}^2(a-b{)}^2$

$L.H.S.=(a^2+b^2+c^2{)}^3-54a^2{b}^2{c}^2$

$=(a^2+b^2+a^2+b^2+2ab{)}^3-54(a^2{b}^2(a+b{)}^2)$ ....... $[\because a+b+c=0]$

$=\left[2\right(a^2+b^2+ab){]}^3-54(a^2{b}^2(a+b{)}^2)$

$=8(a^2+ab+b^2{)}^3-54(a^2{b}^2(a+b{)}^2)$

$=8\left\{\right(a-b{)}^2+3ab{\}}^3-54a^2{b}^2\left\{\right(a-b{)}^2+4ab\}$

$=8(a-b{)}^6+72ab(a-b{)}^4+162a^2{b}^2(a-b{)}^2$

$=2(a-b{)}^2\{2(a-b{)}^2+9ab{\}}^2$

$=2(a-b{)}^2(2a^2+5ab+2b^2{)}^2$

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

$=2(a-b{)}^2(a-c{)}^2(b-c{)}^2=R.H.S.[\because a+b+c=0]$

Hence Proved