<s>[INST] This simple algebra question \begin{cases}
3x + y = \frac{12}{y^2}\\
3y + x = \frac{15}{x^2}
\end{cases}
$x+y = ?$
I tried adding side by side but I found only $4x+4y = \dfrac{15}{x^2} + \dfrac{12}{y^2}$ so I couldn't find anything. Can someone solve it? 
 [/INST] *

*Multiply the first equation by $y^2$: $3xy^2+y^3 = 12$

*Multiply the second equation by $x^2$: $3x^2y+x^3=15$

*Add them up to get $3xy^2+y^3+3x^2y+x^3=(x+y)^3=27.$


Hence, $x+y=3$.
</s>