<s>[INST] Find the value of $x^2+y^2$, given that $x^3-3xy^2=3$ and $y^3-3x^2y=3\sqrt{2}$ Let $x,y$ be reals that satisfy the equations:
$$\begin{align*}
x^3-3xy^2&=3 \\ 
y^3-3x^2y&=3\sqrt{2}
\end{align*}$$
Find the value of $x^2+y^2.$

The terms in the above equations are similar to the expansion $(x\pm y)^3=x^3\pm y^3\pm3xy(x\pm y)$. But subtracting the two equations leads to $$x^3+3x^2y-3xy^2-y^3=(x-y)^3+6xy(x-y)=(x-y)\left((x-y)^2+6xy\right)=3-3\sqrt{2}$$
I also tried solving for $x$ and $y$ using the substitution $y=xt$,
$$\sqrt{2}x^3-3\sqrt{2}x^3t^2=x^3t^3-3x^3t\implies t^3+3\sqrt{2}t^2-3t-\sqrt{2}=0 $$
Both of these approaches don't seem to be particularly useful.
 [/INST] $\begin{array}\\
(x+iy)^3
&=x^3+3ix^2y+3i^2xy^2+i^3y^3\\
&=x^3+3ix^2y-3xy^2-iy^3\\
&=x^3-3xy^2+i(3x^2y-y^3)\\
&=3-3i\sqrt{2}\\
&=3(1-i\sqrt{2})\\
&=3\sqrt{3}(1/\sqrt{3}-i\sqrt{2/3})\\
&=3\sqrt{3}(1/\sqrt{3}-i\sqrt{2/3})\\
&=3\sqrt{3}(\cos(a)-i\sin(a))
\qquad \cos(a)=1/\sqrt{3}\\
&=3\sqrt{3}e^{ia}\\
x+iy
&=\sqrt{3}e^{ia/3}\\
x^2+y^2
&=|x+iy|^2\\
&=3\\
\end{array}
$
In general, if
$x^3-3xy^2=u,
y^3-3x^2y=v
$
then
$\begin{array}\\
(x+iy)^3
&=x^3+3ix^2y+3i^2xy^2+i^3y^3\\
&=x^3+3ix^2y-3xy^2-iy^3\\
&=x^3-3xy^2+i(3x^2y-y^3)\\
&=u-iv\\
(x-iy)^3
&=u+iv\\
(x^2+y^2)^3
&=(x+iy)^3(x-iy)^3\\
&=(u+iv)(u-iv)\\
&=(u^2+v^2)\\
x^2+y^2
&=(u^2+v^2)^{1/3}\\
\end{array}
$
Here
$u=3, v=3\sqrt{2}$.
</s>