Solving equation with complex roots. I have the following question. My problem lies in (c).
Question
a) Find the three roots of the equation $(w+5)(w+8)(w+9)=360$.
b) Let $z_0=\sqrt{-2+6i}$, where $i^2={-1}$. Show that the solutions to the equation $z^2+6z=-11+6i$ can be expressed in the form $z=-3\pm z_0$.
c) Hence, solve the equation $(z+1)(z+2)(z+3)^2(z+4)(z+5)=360$, by reducing it to the equation in (a). Give your answer in terms of $z_0$.
Solution
a) Expanding $(w+5)(w+8)(w+9)=360$ gives, $w^3+22w^2+157w+360=360$.
Cancelling the zeros and taking out the common leaves, $w(w^2+22w+157)=0$ and so, $w=0$ and $w^2+2w+157=0$. The other two roots are
\begin{align*}
w=\dfrac{-22\pm\sqrt{22^2-4(1)(157)}}{2}=-11\pm6i.
\end{align*}
b) Rearranging the equation gives the quadratic $z^2+6z+(11-6i)=0$. Hence,
\begin{align*}
z=\dfrac{-6\pm\sqrt{4(1)(11-6i)}}{2}=-3\pm\sqrt{-2+6i}=-3\pm z_0.
\end{align*}
c)For what it's worth it is clear that $z=0$ is a solution. I tried substituting $-3+z_0$ into the given equation. This produces
\begin{align*}
(z_0^2-4)(z_0^2)(z_0^2-1)=360.
\end{align*}
I do not know what else I can try.

Note that:
$$(z+1)(z+5)=z^2+6z+5$$
$$(z+2)(z+4)=z^2+6z+8$$
$$(z+3)^2=z^2+6z+9$$
So if you put $w=z^2+6z$ you get $(w+5)(w+8)(w+9)=360$ from (a).
