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?

According to Vieta's formulas:
$$\begin{align}a+b+c&=-3\\
ab+bc+ca&=-7\\
abc&=-1\\
a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=9+14=23\\
a^2b^2+b^2c^2+c^2a^2&=(ab+bc+ca)^2-2abc(a+b+c)=49-6=43\\
\end{align}$$
Hence:
$$\begin{align}f'(a)f'(b)f'(c)&=(3a^2+6a-7)(3b^2+6b-7)(3c^2+6c-7)=\\
&=27(abc)^2+54abc(ab+bc+ca)-63(a^2b^2+b^2c^2+c^2a^2)+\\
&\ \ \ \ \ \ 108abc(a+b+c)+216abc-252(ab+bc+ca)+294(a+b+c)-343-\\
&\ \ \ \ \ \ 126(a^2b+ab^2+a^2c+ac^2+b^2c+bc^2)=\\
&=27+54\cdot 7-63\cdot 43+108\cdot 3+147\cdot 23-216+252\cdot 7-294\cdot 3-343-\\
&\ \ \ \ \ \ 126(ab(-3-c)+ac(-3-b)+bc(-3-a))=\\
&=1724-126(-3\cdot (-7)-3\cdot (-1))=\\
&=1724-126\cdot 24=\\
&=1724-3024=\\
&=-1300.\end{align}$$
WA expansion of $(3a^2+6a-7)(3b^2+6b-7)(3c^2+6c-7)$.
