<s>[INST] Find all real values of x and y such that $\left(x+yi\right)^3$ is real and greater than $8$, represent these values in the xy-plane My solution
$\left(x+yi\right)^3>8$
$\left(x^3-3xy^2-8\right)+\left(3x^2y-y^3\right)i>0$
and now I'm stuck with these equations
$\left(x^3-3xy^2-8\right)>0,\left(3x^2y-y^3\right)>0$
 [/INST] I don't know if my answer is correct or not but this is my idea:
$(x+yi)^3= r^3(\cos\phi+i\sin\phi)^3,$
where $r=\sqrt{x^2+y^2}$ and $\phi \in [-\pi,\pi)$.
By de Moivre's formula, we have
$(x+yi)^3= r^3(\cos\phi+i\sin\phi)^3 = r^3(\cos(3\phi)+i\sin(3\phi)).$
Since this is real, then $\sin(3\phi)=0 \Leftrightarrow \phi = \dfrac{k\pi}{3}$, where $k\in \lbrace -1, 0, 1\rbrace$.
Then  $(x+yi)^3= r^3 = \left(\sqrt{x^2+y^2} \right)^3$.
So $ \left(\sqrt{x^2+y^2} \right)^3 >8 \Leftrightarrow x^2+y^2 > 4$. Then in the $xy$-plane, this is the domain outside a circle with $(0,0)$ as a center and radius $2$.
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