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 P(1) = 1, P(2) = 8, P(3) = 27, P(4) = 64, then the value of P(5) is ________.

Clearly, $\mathrm{P}\left(\mathrm{x}\right)-{\mathrm{x}}^3=0$ has roots 1, 2, 3, 4.

$\begin{array}{l}\therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{P}\left(\mathrm{x}\right)-{\mathrm{x}}^3=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)\\ \Rightarrow \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\mathrm{P}\left(\mathrm{x}\right)=(\mathrm{x}-1)(\mathrm{x}-2)(\mathrm{x}-3)(\mathrm{x}-4)+{\mathrm{x}}^3\end{array}$

Hence, $\mathrm{P}\left(5\right)=1\times 2\times 3\times 4+125=149$