Solve $z^3 + 1 = 0$ where $z$ is a complex number I'm working on this problem and I got stuck, here is my attempt.

Find all $z\in\mathbb{C}$ such that $z^3 + 1 = 0$.

Solution. Since $z$ is a complex number, then we must have $z = x + yi$ where $x,y$ are real numbers, hence
$$\left(x + yi\right)^3 = -1$$
or
$$x^3 + 3x^2yi - 3xy^2 - y^3i = -1$$
by equality of the real and imaginary part, we obtain the system 
$$\mbox{$x^3 + 3xy^2 = - 1$ and
$3x^2y - y^3 = 0$}$$
and this is where I got stuck, I even made some substitution to eliminate one variable but it didn't help matters. 
Please someone tell me what to do next. And please don't try solving it using polar forms, just algebraic manipulation. Thanks.

Hint. Note that $z=-1$ is a solution of $z^3+1=0$. It follows that $z^3+1=(z+1)(z^2-z+1)$. Hence, it remains to solve in $\mathbb{C}$ the quadratic equation
$$z^2-z+1=\left(z-\frac{1}{2}\right)^2+\frac{3}{4}=0.$$
Can you take it from here?
