This is equivalent to finding $f(0)$. So, we substitute $x=0$ in the determinant and calculate its value.
\[f(0)=\left|\begin{array}{ccc}0^{3}+1 & 1 & 0 \\ 0^{2}-0 & -1 & 2 \\ 0^{5}+0^{3}+1 & 0 & 1\end{array}\right| = \left|\begin{array}{ccc}1 & 1 & 0 \\ 0 & -1 &
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