Give your answers in the form $x+ iy$, where $x\in \mathbb{R}\:and\:y\in \mathbb{R}$.
I suspect I'll have to use de Moivre's theorem to solve this, but I don't know how to factor in the $+1$, because increasing the real part of a complex number by one doesn't necessarily increase its modulus by one. I know that one of the roots will be $-2$, by inspection, but that's about it.
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Try Solving $y^3 = -1$ to get $y = - e^{2 k \pi /3}$ for $k\in\lbrace 0,1,2 \rbrace$ and then find the three points such that $z = y-1$ for the three values of $y$
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Since you know that $z=-2$ is one of the roots, then it makes sense to expand the left hand side, move the $-1$ over from the right, and factor the resulting cubic:
$$(z+1)^3=-1\iff z^3+3z^2+3z+1=-1\iff z^3+3z^2+3z+2=0\iff(z+2)(z^2+z+1)=0$$
so the other two roots are the roots of the quadratic,
$$z={-1\pm\sqrt{1-4}\over2}={-1\pm\sqrt{-3}\over2}$$
Consider instead $$c^3 = -1$$ where $c = z + 1$. Denoting in geometric form $c = re^{i\theta}$ and $-1 = e^{i \pi}$, we get $$r^3 e^{i3\theta} = e^{i\pi + 2k\pi}, \qquad k \in \lbrace 0,1,2 \rbrace$$ We get $r = 1$ and three different angles \begin{align} \theta_1 &= \frac{\pi}{3} = \\ \theta_2 &= \frac{\pi}{3} + \frac{2\pi}{3} = \pi\\ \theta_3 &= \frac{\pi}{3} + \frac{4\pi}{3} = \frac{5\pi}{3}\\ \end{align} So we get $c_1 = e^{i\frac{\pi}{3}}$, $c_2 = e^{i\pi}$ and $c_3 = e^{i\frac{5\pi}{3}}$. The corresponding roots are \begin{align} z_1 &= -1 + e^{i\frac{\pi}{3}} = (-1 + \cos(\frac{\pi}{3})) + i\sin(\frac{\pi}{3}) \\ z_2 &= -1 + e^{i\pi} = (-1 + \cos(\pi)) + i\sin(\pi)\\ z_3 &= -1 + e^{i\frac{5\pi}{3}}= (-1 + \cos(\frac{5\pi}{3})) + i\sin(\frac{5\pi}{3})\\ \end{align}