<s>[INST] Solving for $x$ and $y$ when $(3x + y)(x + 3y)\sqrt{xy} = 14$ Solve for $x$ and $y$ when $$(3x + y)(x + 3y)\sqrt{xy} = 14$$ $$(x+y)(x^2 + 14xy + y^2) = 36.$$

I was thinking of squaring the first equation and moving on from there, but I think it'll be a bit too messy. Is there a better way to start this problem?
 [/INST] Setting $p=x+y$ and $q=\sqrt{xy}$ (almost as suggested by Alexey in the comments, but $\sqrt{xy}$ looks like it will be more symmetric) and expanding gives
\begin{align}
   3p^2q + 4q^3 &= 14 \\
   p^3 + 12pq^2 &= 36
\end{align}
From here, a lucky coincidence is that
$$
   (p + 2q)^3 = p^3 + 6p^2q + 12pq^2 + 8q^3 = 36 + 2 \cdot 14 = 64
$$
and
$$
   (p - 2q)^3 = p^3 - 6p^2q + 12pq^2 - 8q^3 = 36 - 2 \cdot 14 = 8
$$
which gives us $p+2q = 4$ and $p-2q = 2$. Therefore $p=3$ and $q = \frac12$, giving us $x,y = \frac{3 \pm 2\sqrt2}{2}$.
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