<s>[INST] Find all three complex solutions of the equation $z^3=-10+5i$ Let $z\in \mathbb{C}$. I want to calculate the three solutions of the equation $z^3=-10+5i$. Give the result in cartesian and in exponential representation.
Let $z=x+yi $.
Then we have $$z^2=(x+yi)^2 =x^2+2xyi-y^2=(x^2-y^2)+2xyi$$
And then $$z^3=z^2\cdot z=[(x^2-y^2)+2xyi]\cdot [x+yi ] =(x^3-xy^2)+2x^2yi+(x^2y-y^3)i-2xy^2=(x^3-3xy^2)+(3x^2y-y^3)i$$
So we get
$$z^3=-10+5i \Rightarrow (x^3-3xy^2)+(3x^2y-y^3)i=-10+5i \\ \begin{cases}x^3-3xy^2=-10 \\ 3x^2y-y^3=5\end{cases} \Rightarrow \begin{cases}x(x^2-3y^2)=-10 \\ y(3x^2-y^2)=5\end{cases}$$
Is everything correct so far? How can we calculate $x$ and $y$ ? Or should we do that in an other way?
 [/INST] The exact algebraic answer, which expressible in terms of real quantities by real-valued radicals, is:
$$\bbox[5px,border:2px solid #C0A000]{\begin{align}&x=-\alpha_k\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\\
&y=-\sqrt [3]{\frac{5}{3\alpha_k^2-1}}\end{align}}$$
where,
$$\begin{align}\alpha_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}\arccos\left(-\frac{2\sqrt 5}{5}\right)-\frac{2\pi k}{3}\right)\end{align}$$
for $k\in\{0,1,2\}\thinspace .$

$\rm Construction:$
In this answer, we find all pairs of $(x,y)\in\mathbb R^{2}$ in terms of real quantities by real-valued radicals.
After expanding the parentheses, we have:
$$\begin{align}&\begin{cases}x(x^2-3y^2)=-10 \\y(3x^2-y^2)=5\end{cases}\\
\implies &\frac xy\left(\frac {\frac {x^2}{y^2}-3}{\frac {3x^2}{y^2}-1}\right)=-2\end{align}$$
Substituting $\dfrac xy=u$, leads to:
$$\begin{align}&\frac{u(u^2-3)}{3u^2-1}=-2\\
\implies &u^3+6u^2-3u-2=0\end{align}$$
We have $3$ distinct real roots, however this is the case of Casus irreducibilis.
Using the key substitution $u=v-2$, yields:
$$v^3-15v+20=0$$
We know that, the depressed monic cubic equation $x^3+px+q=0$ is solved by
$$\begin{align}x_k=2\sqrt{-\frac{p}{3}}\cos\left(\frac{1}{3}\arccos\left(\frac{3q}{2p}\sqrt{\frac{-3}{p}}\right)-\frac{2\pi k}{3}\right) \quad \text{for} \quad k=0,1,2.\end{align}$$
Thus we obtain:
$$\begin{align}u_k=-2+2\sqrt{5}\cos\left(\frac{1}{3}\arccos\left(-\frac{2\sqrt 5}{5}\right)-\frac{2\pi k}{3}\right) \quad \text{for} \quad k=0,1,2.\end{align}$$
Finally, setting $\dfrac xy=\alpha\,$ we get:
$$\begin{align}&y(3y^2\alpha^2-y^2)+5=0\\
\implies &(3\alpha^2-1)y^3+5=0\\
\implies &y^3=-\frac{5}{3\alpha^2-1}\\
\implies &y=-\sqrt [3]{\frac{5}{3\alpha^2-1}}\\
\implies &x=-\alpha\sqrt [3]{\frac{5}{3\alpha^2-1}}\end{align}$$
where $\alpha\in\{u_0,u_1,u_2\}\thinspace.$

Note that, casus irreducibilis cannot be solved in radicals in terms of real quantities, it can be solved trigonometrically in terms of real quantities.
Therefore, we can no longer escape the general trigonometric or another non-algebraic representations.
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