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

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(D) Given $f(x) = \left| \begin{array}{ccc} x^3 & x^2 & 3x^2 \ 1 & -6 & 4 \ p & p^2 & p^3 \end{array} ight|$.Expanding the determinant along the f

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$A$ constant

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(D) Given $f(x) = \left| \begin{array}{ccc} x^3 & x^2 & 3x^2 \ 1 & -6 & 4 \ p & p^2 & p^3 \end{array} ight|$.Expanding the determinant along the first row:$f(x) = x^3(-6p^3 - 4p^2) - x^2(p^3 - 4p) + 3x^2(p^2 + 6p)$.Simplifying the expression:$f(x) = (-6p^3 - 4p^2)x^3 + (-p^3 + 4p + 3p^2 + 18p)x^2$.$f(x) = (-6p^3 - 4p^2)x^3 + (-p^3 + 3p^2 + 22p)x^2$.Now,finding the derivatives with respect to $x$:$\frac{df}{dx} = 3(-6p^3 - 4p^2)x^2 + 2(-p^3 + 3p^2 + 22p)x$.$\frac{d^2f}{dx^2} = 6(-6p^3 - 4p^2)x + 2(-p^3 + 3p^2 + 22p)$.$\frac{d^3f}{dx^3} = 6(-6p^3 - 4p^2) = -36p^3 - 24p^2$.Since $p$ is a constant,$-36p^3 - 24p^2$ is also a constant.Therefore,the third derivative is a constant.

$f(x) = \left| \begin{array}{ccc} x^3 & x^2 & 3x^2 \\ 1 & -6 & 4 \\ p & p^2 & p^3 \end{array} \right|$,where $p$ is a constant,then $\frac{d^3f(x)}{dx^3}$ is

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