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On $|z|=3$, the perturbation has modulus $|icz+8|\le 3c+8<11<27=|z^3|$. Thus Rouché's theorem gives three roots inside $|z|<3$, counted with multiplicity.

For the quadrant information, compare instead with $P_0(z)=z^3+8$, whose roots are $r_1=-2$, $r_2=1+i\sqrt{3}$, $r_3=1-i\sqrt{3}$. Their pairwise distances are $2\sqrt{3}$. On the circle $|z-r_j|=1$,The three unit discs are disjoint, and Rouché localization of polynomial roots gives one root in each, counted with multiplicity. The disc around $1+i\sqrt{3}$ lies strictly in the first quadrant in its interior; that around $1-i\sqrt{3}$ lies in the fourth quadrant; and that around $-2$ lies in $\mathrm{Re}z<-1$. Consequently there is exactly one root in each of those two right-hand quadrants, and exactly one root in the left half-plane. The same count also shows that each root is simple.