How to handle indices with fractional degree? An algebra problem ate my head!!! 


  $x$ and $y$ are positive real numbers such that 
  $$\sqrt{x^2 + \sqrt[3]{x^4 y^2}} + \sqrt{y^2 + \sqrt[3]{x^2 y^4}} = 512.$$
  Find $x^{2/3} + y^{2/3}$.


It would be a great help if anybody helps me in solving this problem. I tried taking conjugates and all but I didn't get any answer.
thank you
Hint: Notice that $\sqrt{x^2+\sqrt[3]{x^4y^2}}+\sqrt{y^2+\sqrt[3]{x^2y^3}}$ $= \sqrt{x^2+x^{4/3}y^{2/3}}+\sqrt{y^2+x^{2/3}y^{4/3}}$ 

$= \sqrt{x^{4/3}(x^{2/3}+y^{2/3})}+\sqrt{y^{4/3}(y^{2/3}+x^{2/3})}$ $= x^{2/3}\sqrt{x^{2/3}+y^{2/3}}+y^{2/3}\sqrt{y^{2/3}+x^{2/3}}$ 

$= (x^{2/3}+y^{2/3})\sqrt{x^{2/3}+y^{2/3}}$ $= (x^{2/3}+y^{2/3})^{3/2}$. 

Can you finish the problem from here?
Let, 
\begin{align}
&amp; x^{2/3} + y^{2/3} = z \newline
\implies &amp; \left(x^{2/3} + y^{2/3}\right)^{3/2} = z^{3/2}\newline
\implies &amp; z^{3/2} = 512 = 8^3 \tag{1}
\end{align}

To remove the fractional index $3/2$ on the LHS of (1) to obtain the base $z$, we take root $2/3$ (i.e. two-third root) of both sides, i.e.

\begin{align}
&amp; \left(z^{3/2}\right)^{2/3} = \left(8^3\right)^{2/3} \newline
\implies &amp; z^{(3/2) \times (2/3)} = 8^{3 \times 2/3} \newline
\implies &amp; z = 8^2 = 64
\end{align}

Hence,  $x^{2/3} + y^{2/3} = z = 64$.