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Let $f(x)=\left|\begin{array}{ccc}x^3 & \sin x & \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|$, where $p$ is a constant. Then $\frac{d^3}{d x^3}\{f(x)\}$ at $x=0$ is
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Independent of $p$
$f^{\prime \prime \prime}(x)=\left|\begin{array}{ccc}\frac{d^3}{d x^3} x^3 & \frac{d^3}{d x^3} \sin x & \frac{d^3}{d x^3} \cos x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|=\left|\begin{array}{ccc}6 & -\cos x & \sin x \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|$
$\therefore f^{\prime \prime \prime}(0)=\left|\begin{array}{ccc}6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|=0$, which is independent of p.
$\therefore f^{\prime \prime \prime}(0)=\left|\begin{array}{ccc}6 & -1 & 0 \\ 6 & -1 & 0 \\ p & p^2 & p^3\end{array}\right|=0$, which is independent of p.
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