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
$\left(x^3-3xy^2-8\right)>0,\left(3x^2y-y^3\right)>0$
 A: We need the cube is real then
$$3x^2y-y^3 =y(3x^2-y^2)=0$$
(and not $3x^2y-y^3>0$) then we have three cases

*

*$y=0 \land x\neq 0 \implies x^3-8>0 \implies \begin{cases}x>2\\y=0 \end{cases}$


*$y\neq 0 \land 3x^2=y^2\neq 0 \implies x^3-9x^3-8>0 \implies \begin{cases}x<-1\\y=\pm \sqrt 3 x \end{cases}$


*$y=0 \land x= 0 \implies -8>0$

As an alternative by polar coordinates
$$z=re^{i\theta} \implies z^3=r^3e^{i3\theta}$$
that is
$$r^3\cos (3\theta)>8 \land \sin (3\theta)=0$$
therefore we have

*

*$3\theta = 2k\pi \implies \theta = \frac 2 3 k\pi$, with $k=0,1,2$

*$r>2$
which indicates that solutions are three oriented lines out of the circle $x^2+y^2=4$.
A: 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 $2$.
