Originally Posted by
harish21 Let $\displaystyle u = x^{-1/3} $.
then the given equation becomes:
$\displaystyle \frac{2u}{3}-\frac{u^2}{3}=0$
$\displaystyle \frac{2u - u^2}{3} = 0$
$\displaystyle 2u - u^{2} = 0$
$\displaystyle u^{2} - 2u = 0$
$\displaystyle u(u-2) = 0$
So, $\displaystyle u = 0$ OR $\displaystyle u = 2$
substitute $\displaystyle u= x^{\frac{-1}{3}}$, you get
$\displaystyle x^{\frac{-1}{3}} = 2$
cube both sides to get:
$\displaystyle x^{-1} = 8$
that is
$\displaystyle \frac{1}{x} = 8$
therefore,
$\displaystyle x = \frac{1}{8}$