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

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    $\begingroup$ I edited your post to include mathjax formatting. The picture was quite unclear, so can you please check that my edit is correct? $\endgroup$ – user296602 Dec 24 '15 at 7:39
  • $\begingroup$ ^The picture is very unclear, but I think it asks to compute $x^{\color{red}{2/3}}+y^{2/3}$ and not $x^{\color{red}{3/4}}+y^{2/3}$. $\endgroup$ – JimmyK4542 Dec 24 '15 at 7:58
  • $\begingroup$ @JimmyK4542 I think you're right. I (hopefully) fixed the original according to that. $\endgroup$ – user296602 Dec 24 '15 at 7:59
  • $\begingroup$ yes jimmys right $\endgroup$ – user300518 Dec 24 '15 at 8:09

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?

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  • $\begingroup$ thank you very much sir. i got the answer. but i want to know the thinking process. knowing the answer wont help me in the exam, right? sir please tell me the thinking process of yours. how did it come to your mind to do this? $\endgroup$ – user300518 Dec 24 '15 at 8:13
  • $\begingroup$ Since you only have one equation and two variables, it is likely that the left side simplifies to something in terms of $x^{2/3}+y^{2/3}$ only. So to simplify the left side, I tried pulling the "greatest common factor" out of the square roots, and that immediately led to the factorization. The process might be easier to see if you substitute $x = a^3$ and $y = b^3$. Then all the exponents become nice integers. $\endgroup$ – JimmyK4542 Dec 24 '15 at 8:22

Let, \begin{align} & x^{2/3} + y^{2/3} = z \newline \implies & \left(x^{2/3} + y^{2/3}\right)^{3/2} = z^{3/2}\newline \implies & 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} & \left(z^{3/2}\right)^{2/3} = \left(8^3\right)^{2/3} \newline \implies & z^{(3/2) \times (2/3)} = 8^{3 \times 2/3} \newline \implies & z = 8^2 = 64 \end{align}

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

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