If $\mathrm{f}\left(\mathrm{x}\right)=\left|\begin{array}{ccc}\mathrm{x}-3& 2{\mathrm{x}}^2-18& 3{\mathrm{x}}^3-81\\ \mathrm{x}-5& 2{\mathrm{x}}^2-50& 4{\mathrm{x}}^3-500\\ 1& 2& 3\end{array}\right|$ ,then $\mathrm{f}\left(1\right)\cdot \mathrm{f}\left(3\right)+\mathrm{f}\left(3\right)\cdot \mathrm{f}\left(5\right)\cdot \mathrm{f}\left(5\right)\cdot \mathrm{f}\left(1\right)$ is equal to

$\begin{array}{l}\mathrm{f}\left(1\right)=\left|\begin{array}{rrr}-2& -16& -78\\ -4& -48& -496\\ 1& 2& 3\end{array}\right|\\ \mathrm{f}\left(3\right)=\left|\begin{array}{rrr}0& 0& 0\\ -2& -32& -392\\ 1& 2& 3\end{array}\right|=0\end{array}$

$\begin{array}{c}f\left(1\right)f\left(3\right)+f\left(3\right)f\left(5\right)f\left(5\right)f\left(1\right)=f\left(1\right)f\left(3\right)\left[1+\left\{f\right(5){\}}^2\right]\\ =0\times \left[1+\left\{f\right(5){\}}^2\right]\\ =0\end{array}$