Suppose  $P\left(x\right)$ is a cubic polynomial with integer coefficients such that $P\left(\sqrt{5}\right)=5$ and  $P\left(\sqrt[3]{5}\right)=5\sqrt[3]{5}$. Compute  $\left|P\left(5\right)\right|$.

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