Hello can you please help me to solve this equation.
$\displaystyle z^3+8=0$
An alternative...
$\displaystyle \displaystyle \begin{align*} z^3 + 8 &= 0 \\ z^3 &= -8 \\ z^3 &= 8e^{i\pi} \\ z &= \left(8e^{i\pi}\right)^{\frac{1}{3}} \\ z &= 2e^{\frac{i\pi}{3}} \end{align*}$
Since this is a cubic, there are three solutions, all evenly spaced around a circle. So their angles are separated by $\displaystyle \displaystyle \frac{2\pi}{3}$, which means the three solutions are $\displaystyle \displaystyle 2e^{-\frac{i\pi}{3}}, 2e^{\frac{i\pi}{3}}$ and $\displaystyle \displaystyle 2e^{i\pi}$.
Convert back to Cartesians.