Cube root of 1

$$ The cube root of 1 can be obtained by solving $z^3=1$ . Therefore $$ $\begin{aligned} z^3&=1\\ z^3-1&=0\\ (z-1)(z^2 + z + 1) &= 0 \end{aligned}$ $$ Clearly that solution of the last equation has 3 roots. One real root from the solution of $z-1=0$ and two complex roots from the solution of $z^2 + z + 1=0$ . Thus, the cube roots of 1 are: $$ $\sqrt[3]{1} =\left\{ \begin{array}{l l} \quad 1,\\ \\ -\frac{1}{2}+\frac{1}{2}\sqrt{3}\;i,\\ \\ -\frac{1}{2}-\frac{1}{2}\sqrt{3}\;i. \end{array} \right.$ $$ $\text{\# }\mathbb{Q.E.D.}\text{ \#}$