Solving imaginary equation $z^3 = 5i + 5$

$$z^3=5(1+i)$$

$$=5\sqrt{2}(\frac{1}{\sqrt{2}}+\frac{i}{\sqrt{2}})$$

$$=5\sqrt{2}(\cos(\frac{\pi}{4})+i\sin(\frac{\pi}{4}))$$

Do you now see some thing worthy???

Alternatively, you can write $$5+5i=\sqrt{50}\operatorname{cis}\frac\pi4,$$ so putting $$\zeta=\sqrt[6]{50} \operatorname{cis} \frac\pi{12},$$ we have $$z^3=\zeta^3\\z^3-\zeta^3=0\\(z-\zeta)(z^2+\zeta z+\zeta^2)=0,$$ and so the solutions are $z=\zeta$, and the solutions to the quadratic equation $z^2+\zeta z+\zeta^2=0.$ As yet another alternative, put $\omega= \operatorname{cis} \frac{2\pi}3,$ so that $1,\omega,\omega^2$ are the complex cube roots of $1$, and so $\zeta,\zeta\omega,\zeta\omega^2$ are the solutions to the equation $z^3=\zeta^3.$

Regardless you should find solutions $$z=\sqrt[6]{50} \operatorname{cis}\left(\frac\pi{12}+\frac{2\pi k}3\right),$$ for $k=0,1,2.$