Solution:

Write $P(x)=a x^{3}+b x^{2}+c x+d$, where $a, b, c, d$ are integers. Then we have that
$$\begin{aligned} P(\sqrt{5})-5 & =(5 a+c) \sqrt{5}+(5 b+d-5)=0 \\ P(\sqrt[3]{5})-5 \sqrt[3]{5} & =(5 a+d)+(c-5) \sqrt[3]{5}+b \sqrt[3]{25}=0 \end{aligned}$$
Recall that $\sqrt{5}$ is irrational. In particular, since $(5 a+c) \sqrt{5}+(5 b+d-5)=0$, we must have $5 a+c=0$ and $5 b+d-5=0$. Similarly, from the condition on $\sqrt[3]{5}$, we must have $5 a+d=c-5=b=0$.
This is enough to imply $(a, b, c, d)=(-1,0,5,5)$, so $P(x)=-x^{3}+5 x+5$. Hence, our final answer is $P(5)=-125+25+5=-95$.