Let $\mathrm{P}\left(\mathrm{x}\right)={\mathrm{x}}^4+{\mathrm{ax}}^3+{\mathrm{bx}}^2+\mathrm{cx}+\mathrm{d}$be a polynomial such that $\mathrm{P}\left(1\right)=1,\mathrm{P}\left(2\right)=8,\mathrm{P}\left(3\right)=27,\mathrm{P}\left(4\right)=64$,then the value of P(5) is_________.

Clearly, $\mathrm{P}\left(\mathrm{x}\right)-{\mathrm{x}}^3=0\text{ has roots }1,2,3,4$
$\begin{array}{r}\therefore \mathrm{P}\left(\mathrm{x}\right)-{\mathrm{x}}^3=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)\\ \Rightarrow \mathrm{P}\left(\mathrm{x}\right)=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)+{\mathrm{x}}^3\end{array}$
Hence, $\mathrm{P}\left(5\right)=1\times 2\times 3\times 4+125=149$.