I just wanted to check if anyone can see a simpler way to solve this. Because I am not looking forward to using the cubic formula to solve it!
$$ det(\lambda-AI) = \left| \begin{array}{ccc}
\lambda + a & -b & -b \\
c & \lambda + d & 0 \\
0 & d & \lambda \end{array} \right| = 0$$
requires us to solve
$$ \lambda^3 + (a+d)\lambda^2 + (ad+bc)\lambda -bcd = 0 $$
(I can find no rational roots. As, by the rational zero theorem, the rational roots would be $\pm$ a factor of bcd. The only factors which will cancel out bcd are $\lambda = \pm\sqrt{bc}$ or $\lambda = -d$ neither of which result in $ det(\lambda-AI) = 0$.)
Is the next step the cubic formula? :(
Just to clarify for those who may see this question in the future, the actual matrix was
$$A = \begin{pmatrix} -a & -b & -b\\ c & -d & 0\\ 0 & d & 0 \end{pmatrix}$$
so that the polynomial to solve was $$\lambda^3 + (a+d)\lambda^2 + (ad + bc)\lambda + bcd,$$
which has an obvious root $\lambda = -d$, and hence factors out.