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

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(C) We have,$f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \ x - a & x^2 - x & c + x \ x - b & x - c & 0 \end{arr

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$f(0) = 0$

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(C) We have,$f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \ x - a & x^2 - x & c + x \ x - b & x - c & 0 \end{array} ight|$.
To find $f(0)$,we substitute $x = 0$ into the determinant:
$f(0) = \left| \begin{array}{ccc} 0^3 - 0 & a + 0 & b + 0 \ 0 - a & 0^2 - 0 & c + 0 \ 0 - b & 0 - c & 0 \end{array} ight| = \left| \begin{array}{ccc} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{array} ight|$.
Let $A = \left[ \begin{array}{ccc} 0 & a & b \ -a & 0 & c \ -b & -c & 0 \end{array} ight]$.
Since $A^T = \left[ \begin{array}{ccc} 0 & -a & -b \ a & 0 & -c \ b & c & 0 \end{array} ight] = -A$,the matrix $A$ is a skew-symmetric matrix of order $3$.
The determinant of a skew-symmetric matrix of odd order is always $0$.
Therefore,$f(0) = 0$.

If $f(x) = \left| \begin{array}{ccc} x^3 - x & a + x & b + x \\ x - a & x^2 - x & c + x \\ x - b & x - c & 0 \end{array} \right|$,then:

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