Factor a cubic and transform its roots

The real root is $w=-2$. Write $w^3+8=(w+2)(w^2+Aw+B)$. Comparing constant terms gives $B=4$ and comparing the coefficient of $w^2$ gives $A=-2$. Thus $(w+2)(w^2-2w+4)=0$. The quadratic formula gives $w=(2\pm\sqrt{4-16})/2=1\pm i\sqrt3$.

Write $w^4+w^3+8w+8=(w^3+8)(w+C)$. Constant terms give $C=1$, so the roots are $-1,-2,1+i\sqrt3,1-i\sqrt3$.

Let $w=iz$. Then $(iz)^4+(iz)^3+8iz+8=z^4-iz^3+8iz+8$. Therefore divide each w-root by i: $iz=-1$ gives $z=i$; $iz=-2$ gives $z=2i$; $iz=1+i\sqrt3$ gives $z=\sqrt3-i$; $iz=1-i\sqrt3$ gives $z=-\sqrt3-i$.