If f(x)=left|begin{array}{ccc}x-3 & 2x^2-18 & | vedclass

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(B) Given $f(x) = \left|\begin{array}{ccc} x-3 & 2(x^2-9) & 2x^3-81 \ x-5 & 2(x^2-25) & 4(x^3-125) \ 1 & 2 & 3 \end{array}ight|$.For $x=5$,the sec

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$2183328$

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(B) Given $f(x) = \left|\begin{array}{ccc} x-3 & 2(x^2-9) & 2x^3-81 \ x-5 & 2(x^2-25) & 4(x^3-125) \ 1 & 2 & 3 \end{array}ight|$.For $x=5$,the second row becomes $5-5=0$,$2(25-25)=0$,and $4(125-125)=0$. Since a row is zero,$f(5)=0$.Now calculate $f(1)$:$f(1) = \left|\begin{array}{ccc} -2 & -16 & -79 \ -4 & -48 & -496 \ 1 & 2 & 3 \end{array}ight| = -2888$.Now calculate $f(3)$:$f(3) = \left|\begin{array}{ccc} 0 & 0 & -27 \ -2 & -32 & -392 \ 1 & 2 & 3 \end{array}ight| = -27(0 - (-32 + 32)) = 756$ (Correction: Calculating determinant: $-27(-4 - (-32)) = -27(28) = -756$).Thus,$f(1) \cdot f(3) + f(3) \cdot f(5) + f(5) \cdot f(1) = (-2888 	imes -756) + 0 + 0 = 2183328$.

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

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