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

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

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$f(3)$

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(B) Given $f(x) = \left| \begin{array}{ccc} x - 3 & 2(x^2 - 9) & 3(x^3 - 27) \ x - 5 & 2(x^2 - 25) & 4(x^3 - 125) \ 1 & 2 & 3 \end{array} ight|$.Observe that if $x = 3$,the first row becomes $(3-3, 2(9-9), 3(27-27)) = (0, 0, 0)$. Since one row is zero,$f(3) = 0$.Observe that if $x = 5$,the second row becomes $(5-5, 2(25-25), 4(125-125)) = (0, 0, 0)$. Since one row is zero,$f(5) = 0$.We need to calculate $f(1)f(3) + f(3)f(5) + f(5)f(1)$.Substituting the values $f(3) = 0$ and $f(5) = 0$ into the expression:$f(1) 	imes 0 + 0 	imes 0 + 0 	imes f(1) = 0 + 0 + 0 = 0$.Since $f(3) = 0$,the result 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) = $

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