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

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(B) Given $f(x) = \left| \begin{array}{ccc} x-3 & 2(x^2-9) & 3(x^3-27) \ x-5 & 2(x^2-25) & 4(x^3-125) \ 1 & 2 & 3 \end{array} ight|$.We can factor

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$f(3)$

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(B) Given $f(x) = \left| \begin{array}{ccc} x-3 & 2(x^2-9) & 3(x^3-27) \ x-5 & 2(x^2-25) & 4(x^3-125) \ 1 & 2 & 3 \end{array} ight|$.We can factor out $(x-3)$ from the first row and $(x-5)$ from the second row:$f(x) = (x-3)(x-5) \left| \begin{array}{ccc} 1 & 2(x+3) & 3(x^2+3x+9) \ 1 & 2(x+5) & 4(x^2+5x+25) \ 1 & 2 & 3 \end{array} ight|$.From the expression,it is clear that $f(3) = 0$ and $f(5) = 0$.Substituting these values into the expression $f(1)f(3) + f(3)f(5) + f(5)f(1)$:$f(1)(0) + (0)(0) + (0)f(1) = 0$.Since $f(3) = 0$,the value of the expression is $f(3)$.

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

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