$f\left( x \right) = x\left( {x + 1} \right)\left( {x - 1} \right)\left|{\begin{array}{{}{c}}1&1&1\\{2x}&{x - 1}&x\\{3x}&{x - 2}&x\end{array}} \right|$ $\left(\because {{C_2}\left( {\frac{1}{x}} \right),{C_3}\left( {\frac{1}{{x - 1}}} \right),{R_3}\left( {\frac{1}{{x - 1}}} \right)} \right)$
$\therefore f\left( x \right) = x\left( {x + 1} \right)\left( {x - 1} \right)\left| {\begin{array}{{}{c}}0&0&1\\{x + 1}&{ - 1}&x\\{2\left( {x + 1} \right)}&{ - 2}&x\end{array}} \right|$ $(\because {C_1} \to {C_1} - {C_2},{C_2} \to {C_2} - {C_3})$
$\therefore f\left( x \right) = - x{\left( {x + 1} \right)^2}\left( {x - 1} \right)\left| {\begin{array}{{}{c}}0&0&1\\1&1&x\\2&2&x\end{array}} \right|$ $\left( \because{{C_1}\left( {\frac{1}{{x + 1}}} \right),{C_2}\left( { - 1} \right)} \right)$
$\therefore f(x)=0$ $\left(\because {{C_1} = {C_2}} \right)$
$\therefore f(100)=0$
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