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?