Question

Show that $x^{a^2{b}^{-1}{c}^{-1}}\cdot x^{a^{-1}{b}^2{c}^{-1}}\cdot x^{a^{-1}{b}^{-1}{c}^2}=x^3$, if $a+b+c=0$.

Answer 1

When $a+b+c=0$

$a+b=-c$

Cubing , we get

$a^3+b^3+3ab(a+b)=-c^3$

$a^3+b^3+c^3=-3ab(a+b)$

$a^3+b^3+c^3=(-3ab)(-c)$

$a^3+b^3+c^3=3abc$

$x^{a^2{b}^{-1}{c}^{-1}}.x^{a^{-1}{b}^2{c}^{-1}}.x^{a^{-1}{b}^{-1}{c}^2}=x^{\frac{a^2}{bc}+\frac{b^2}{ca}+\frac{c^2}{ab}}=x^{\frac{a^3+b^3+c^3}{abc}}=x^{\frac{3abc}{abc}}=x^3$

Hence proved.