Explanation

(d)
$f(x) = \left| \begin{matrix}
x^{3} & x^{2} & 3x^{2} \\
1 & - 6 & 4 \\
p & p^{2} & p^{3}
\end{matrix} \right|$
$\Rightarrow f(x) = x^{3}\left( - 6p^{3} - 4p^{2} \right) - x^{2}\left( p^{3} - 4p \right) + 3x^{2}\left( p^{2} + 6p \right)$
$\Rightarrow f(x) = - 6p^{3}x^{3} - 4p^{2}x^{3} - x^{2}p^{3} + 4px^{2} + 3p^{2}x^{2} + 18px^{2}$
On differentiating w.r.t. $x$, we get
$\frac{d}{dx}f(x) = - 18p^{3}x^{2} - 12p^{2}x^{2} - 2xp^{3} + 8px + 6p^{2}x + 36px$
Again differentiating w.r.t. $x$, we get
$\frac{d^{2}}{dx^{2}}f(x) = - 36p^{3}x^{\ } - 24p^{2}x^{\ } - 2p^{3} + 8p + 6p^{2} + 36p$
Again differentiating w.r.t. $x$, we get
$\frac{d^{3}}{dx^{3}}f(x) = - 36p^{3} - 24p^{2} =$constant