<s>[INST] Efficiently compute $f'(a)f'(b)f'(c)$ for the roots of a cubic polynomial Let $a,b, c$ be the zeroes of $f(x) = x^3+3x^2-7x+1$. Find $f'(a)f'(b)f'(c)$.
My idea involved substituting in $a,b,c$ into $f'(x)$ then using Vieta's but that would take far too long of a time, especially since I'm training for competitions. What would be a great way to solve this within a time limit?
 [/INST] If $f(x)=(x-r)g(x)$, then $f'(x)=g(x)+(x-r)g'(x)$ and hence $f'(r)=g(r)$.
As $f(x)=(x-a)(x-b)(x-c)$, $f'(a)f'(b)f'(c)=(a-b)(a-c)(b-a)(b-c)(c-a)(c-b)$.
Let $g(x)=f(x-1)$ and $\alpha$, $\beta$ and $\gamma$ be the roots of $g(x)=0$. Then $\alpha=a+1$, $\beta=b+1$ and $\gamma=c+1$. Therefore, $g'(\alpha)g'(\beta)g'(\gamma)=f'(a)f'(b)f'(c)$.
$g(x)=(x-1)^3+3(x-1)^2-7(x-1)+1=x^3-10x+10$.
$g'(x)=3x^2-10$.
\begin{align*}
g'(\alpha)g'(\beta)g'(\gamma)&=
\frac{(3\alpha^3-10\alpha)(3\beta^3-10\beta)(3\gamma^3-10\gamma)}{\alpha\beta\gamma}\\
&=\frac{(20\alpha-30)(20\beta-30)(20\gamma-30)}{\alpha\beta\gamma}\\
&=\frac{-8000(\frac{3}{2}-\alpha)(\frac{3}{2}-\beta)(\frac{3}{2}-\gamma)}{\alpha\beta\gamma}\\
&=\frac{-8000}{\alpha\beta\gamma}g\left(\frac{3}{2}\right)\\
&=\frac{-8000}{-10}\left[\left(\frac{3}{2}\right)^3-10\left(\frac{3}{2}\right)+10\right]\\
&=-1300
\end{align*}
We can also use the discriminant of $g(x)=0$, which is $-4(-10)^3-27(10)^2=1300$.
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