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?
 A: Some ideas that are too long for a comment (maybe even a partial answer):
Since the coefficient of $x^3$ is $1$, we have
$$
f(x) = (x-a)(x-b)(x-c)
$$
Differentiating this using the product rule leads us to
$$
f'(x) = (x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b)
$$
So the derivatives asked for are given by
$$
f'(a) = (a-b)(a-c) = a^2 -ab-ac + bc\\
f'(b) = (b-a)(b-c) = b^2 -ab + ac - bc\\
f'(c) = (c-a)(c-b) = c^2 +ab-ac -bc
$$
If you know the roots, then this is the expression I would use to insert and calculate.
If not, then... The sum of these derivatives is, by Vieta's formulas
$$
a^2 + b^2 + c^2 -ab-ac-bc = (a+b+c)^2-3(ab+ac+bc) = (-1)^2 -3\cdot(-7) = 22
$$
and their product is $-1300$, as it's the negative of the discriminant. I am failing to find a third equation to help solve this, though.

Edit: The question has changed to just finding the product, and as I mentioned, that is $-1300$. I'll elaborate a bit on that, then.
Given all the (not necessarily distinct) roots $r_1, r_2, \ldots, r_n$ of a degree-$n$ monic polynomial, the discriminant of that polynomial is
$$
\prod_{i<j}(r_i-r_j)^2
$$
and we see from the expressions above that
$$
f'(a)f'(b)f'(c) = (a-b)(b-a)(a-c)(c-a)(b-c)(c-b)
$$
which is the negative of the discriminant. So how would you know that the discriminant is $1300$ in the first place? Well, that's just a formula to learn in this case. For a quadratic polynomial $rx^2 + sx + t$, the discriminant is $s^2 - 4rt$, while for a cubic polynomial $px^3 + qx^2 + rx + s$ the discriminant is
$$
q^2r^2 - 4q^3s - 4pr^3 - 27p^2s^2 + 18pqrs
$$
In our case, this becomes
$$
9\cdot 49 -4\cdot 27\cdot 1 - 4\cdot1\cdot (-343)-27\cdot 1\cdot 1 + 18\cdot 1\cdot3\cdot(-7)\cdot1 = 1300
$$
A: 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 $-4(-10)^3-27(10)^2=1300$.
A: The answer of @Arthur uses the discriminant of polynomial $f$  (generalizing usual $\Delta$ for quadratic equations) ; The concept of discriminant is a particular case of the resultant, a very powerful concept. Here is a way to work with it.
For the theory of the resultant see this article.
Let us refer to it for formulas involving the resultant of two polynomials $A=a_0x^d+...$ and $B=b_0x^{e}+...$.
$$res(A,B)=(-1)^{de}a_0^{e}\prod_i B(\lambda_i)\tag{1}$$
where the $\lambda_i$s are the roots of polynomial $A$ (it is not an error : we compute values taken by a polynomial on the roots of the other polynomial).
(1) can also be written : 
$$\prod_i  B(\lambda_i)=(-1)^{de}a_0^{e}res(A,B)\tag{2}$$
where the LHS is nothing else than $f'(a)f'(b)f'(c)$ if we take 
$$\begin{cases}A&=&f&=&x^3+3x^2-7x+1 \ \text{and} \\B&=&f'&=&3x^2+6x-7\end{cases}.$$
Using the determinant formula for the resultant 
$$ res(A,B)=\det\begin{pmatrix} 1  &  3  & -7  &  1  &  0\\
     0  &  1  &  3 & -7    & 1\\
     3  &  6  & -7 &   0   & 0\\
     0  &  3  &  6 &  -7   & 0\\
     0  &  0  &  3 &   6   &-7\end{pmatrix}=1300.\tag{3}$$
(Please note that we have written the coefficients of polynomial $A$, resp. $B$, twice, resp. 3 times).
We get the final result 

$$f'(a)f'(b)f'(c)=-1300,$$

due to multiplication by $(-1)^{de}=(-1)^{3}=-1$ in formula (2).
A: $$f(x) = x^3+3x^2-7x+1=(x-a)(x-b)(x-c)$$
$$f'(x)=(x-a) (x-b)+(x-a) (x-c)+(x-b) (x-c)$$
$$f'(a)=(a-b) (a-c)$$
$$f'(b)=(b-a) (b-c)$$
$$f'(c)=(c-a) (c-b)$$
A: 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 $(3a^2+6a-7)(3b^2+6b-7)(3c^2+6c-7)$.
