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?
 A: 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 $x,y = \frac{3 \pm 2\sqrt2}{2}$.
A: From the first equation we have $xy>0$.
Thus, the second give $x+y>0,$ which says $x>0$ and $y>0$.
Let $y=t^2x,$ where $t>0$.
Thus, $$18(3t^2+1)(t^2+3)t=7(t^2+1)(t^4+14t^2+1)$$ or
$$(t^2-6t+1)(7t^4-12t^3+26t^2-12t+7)=0$$ or
$$t^2-6t+1=0,$$
which gives $y=(17-12\sqrt2)x$ or $y=(17+12\sqrt2)x.$
Let $y=(17-12\sqrt2)x.$
Thus, the first equation gives:
$$(3+10(17-12\sqrt2)+3(17-12\sqrt2)^2)(3-2\sqrt2)x^3=14$$ or
$$x^3=\frac{99+70\sqrt2}{8}$$ or
$$x=\frac{1}{2}(3+2\sqrt2)$$ and
$$y=(3-2\sqrt2)^2\cdot \frac{1}{2}(3+2\sqrt2),$$
which gives $$\left(\frac{3+2\sqrt2}{2},\frac{3-2\sqrt2}{2}\right).$$
The second case by the same way gives:
$$\left(\frac{3-2\sqrt2}{2},\frac{3+2\sqrt2}{2}\right).$$
