Vieta's Derivatives

We have $f(x)=x^{3}-x^{2}-2x+1 = (x - \alpha)(x - \beta)(x -\gamma)$

$\Rightarrow f'(x) = (x - \alpha)(x - \beta) + (x - \beta)(x -\gamma) + (x - \alpha)(x -\gamma)$

So ,from this we may calculate $f'(\alpha) , f'(\beta)$ , and $f'(\gamma)$ and get

$f'(\alpha)\times f'(\beta)\times f'(\gamma) = -(\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2$

Now, using the definition of discriminant we have

$\Delta = (\alpha - \beta)^2(\beta - \gamma)^2(\gamma - \alpha)^2$

The discriminant for the cubic polynomial $px^3 + qx^2 + rx + s$ is given by $\Delta = q^2r^2 -4pr^3 -4q^3s - 27p^2s^2 + 18pqrs$

On calculation we get $\Delta = 49$ , so our final answer is $- 49$ .

Here's a non-discriminant approach but without bashing out the solution.

We have $f'(x) = 3x^2 - 2x - 2$ , so we want to find $\displaystyle \prod_{ r = \{ \alpha, \beta, \gamma \} } (3r^2 - 2r - 2)$

Expanding it and comparing it by Vieta's formula will be extremely tedious. So I consider factoring the quadratic polynomial.

Consider $3x^2 - 2x - 2 = 3(x-A)(x-B)$ , solving the identity equation for constants $A,B$ (values are interexchangeable) gives $A = \frac { 1 + \sqrt7}{3},B = \frac { 1 - \sqrt7}{3}$

Thus the desired product becomes

$\begin{aligned} & & 3^3 \displaystyle \prod_{ r = \{ \alpha, \beta, \gamma \} } \left [ \left (r - \frac 1 3 + \frac { \sqrt{7} }{3} \right ) \left (r - \frac 1 3 - \frac { \sqrt{7} }{3} \right ) \right ] \\ & = & 27 \displaystyle \prod_{ r = \{ \alpha, \beta, \gamma \} } \left [ \left (r-\frac 1 3 \right )^2 - \frac {7}{9} \right ] \\ \end{aligned}$

Let $g(x) = f \left (x+ \frac 1 3 \right ) =x^3- \frac 7 3 x+ \frac 7 {27}$ be a cubic function with roots $\alpha^* = \alpha - \frac 13 , \beta^* = \beta - \frac 13 , \gamma^* = \gamma - \frac 13$ . The product further simplifies to

$27 \displaystyle \prod_{ r = \{ \alpha^*, \beta^*, \gamma^* \} } \left [ r^2 - \frac {7}{9} \right ]$

Now consider $h(x) = g ( \sqrt{x} ) = 0$ be a cubic equation has roots $(\alpha^*)^2, (\beta^*)^2, (\gamma^*)^2$

$\begin{aligned} x \sqrt x - \frac 7 3 \sqrt x + \frac {7}{27} & = & 0 \\ x \left (x-\frac 7 3 \right )^2 & = & \left (\frac {7}{27} \right )^2 \\ x^3- \frac {14}{3} x^2+\frac {49}{9} x -\frac {49}{729} & = & 0 \\ \end{aligned}$

So $h(x) = x^3- \frac {14}{3} x^2+\frac {49}{9} x -\frac {49}{729}$ . Product becomes

$27 \displaystyle \prod_{ r = \{ (\alpha^*)^2, (\beta^*)^2, (\gamma^*)^2 \} } \left [ r - \frac {7}{9} \right ]$

So the product above is simply $-27$ (we multiply by $-1$ because by Vieta's formula it's the constant term in a cubic equation) times by the coefficient of the constant term in the expansion of $h \left ( x + \frac {7}{9} \right )$ .

Hence, the answer is $-27 \times \left [ \left ( \frac {7}{9} \right )^3 - \frac {14}{3} \left ( \frac {7}{9} \right )^2 + \frac {49}{9} \left ( \frac {7}{9} \right ) - \frac {49}{729} \right ] = \boxed{-49}$

This just shows that this problem could be solved without expanding the product and without the knowledge of the discriminant of the cubic equation, but doing so will look extremely ridiculous.

$f(x) = x^3 - x^2 - 2x + 1 = 0 \Rightarrow 1 = 2x+x^2-x^3$

Hence, $f'(x) = 3x^2 - 2x - 2(2x+x^2-x^3) = x(2x-3)(x+2)$ .

Let the solutions for $f(x) = 0$ be $a$ , $b$ , and $c$ .

We know that $abc = -1, ab+bc+ca = -2, a+b+c = 1$ .

Therefore,

$f'(a)f'(b)f'(c) = abc[(2a-3)(2b-3)(2c-3)][(a+2)(b+2)(c+2)]$

$= abc[8abc-12(ab+bc+ca)+18(a+b+c)-27][abc+2(ac+bc+ca)+4(a+b+c)+8]$

$= (-1)(7)(7) = -49$

Used TI-83 Plus, graphed the curve, found the roots and calculated the answer from the calculator.

Let $M = \begin{pmatrix} 0 & 0 & -1 \\ 1 & 0 & 2 \\ 0 & 1 & 1 \end{pmatrix}$ the companion matrix of the polynomial $f$ . The matrix $M$ is similar to a triangular one with $\alpha, \beta, \gamma$ on its diagonal, so that for any polynomial $g$ , the matrix $g(M)$ is similar to a triangular one with $g(\alpha), g(\beta), g(\gamma)$ on its diagonal. Hence, the answer is the determinant of the matrix $f'(M) = 3M^2 - 2M - 2 I_3 = \begin{pmatrix} -2 & -3 & -1 \\ -2 & 4 & -1 \\ 3 & 1 & 5 \end{pmatrix}$ .

For easy typing purposes, I will let $\alpha = a = x_1$ , $\beta = b = x_2$ and $\gamma = c = x_3$

$f(x) = x^3 - x^2 - 2x + 1$

$f'(x) = 3x^2 - 2x - 2$

$f'(\alpha) \times f'(\beta) \times f'(\gamma)$

$= (3x_1^2 - 2x_1 - 2)(3x_2^2 - 2x_2 - 2)(3x_3^2 - 2x_3 - 2)$

$= \displaystyle \prod_{i=1}^3 (3x_i^2 - 2x_i - 2)$

The rest of the problem is to find this product, which is exactly the same as this question

From Vieta's formula, we get:

$a + b + c = 1$

$ab + ac + bc = -2$

$abc = -1$

Next, we simplify the product above by obtaining $(x_i^3 - x_i^2)$ in it:

$\displaystyle \prod_{i=1}^3 (3x_i^2 - 2x_i - 2)$

$= \displaystyle \prod_{i=1}^3 \dfrac{(3x_i^2 - 2x_i - 2)(x_i - \dfrac{1}{3})}{x_i - \dfrac{1}{3}}$

$= \displaystyle \prod_{i=1}^3 \dfrac{3x_i^3 - 3x_i^2 - \dfrac{4}{3}x_i + \dfrac{2}{3}}{x_i - \dfrac{1}{3}}$

$= \displaystyle \prod_{i=1}^3 \dfrac{3(x_i^3 - x_i^2) - \dfrac{4}{3}x_i + \dfrac{2}{3}}{x_i - \dfrac{1}{3}}$

Given that $x_i$ are the zeroes of the function, we get:

$x_i^3 - x_i^2 - 2x_i + 1 = 0$

$x_i^3 - x_i^2 = 2x_i - 1$

Substitute this into the product, and we get:

$\displaystyle \prod_{i=1}^3 \dfrac{3(2x_i - 1) - \dfrac{4}{3}x_i + \dfrac{2}{3}}{x_i - \dfrac{1}{3}}$

$= \displaystyle \prod_{i=1}^3 \dfrac{\dfrac{14}{3}x_i - \dfrac{7}{3}}{x_i - \dfrac{1}{3}}$

$= \displaystyle \prod_{i=1}^3 \dfrac{7(2x_i - 1)}{3x_i - 1}$

$= \dfrac{7(2a - 1)(7)(2b - 1)(7)(2c - 1)}{(3a - 1)(3b - 1)(3c - 1)}$

$= 7^3\left(\dfrac{8abc - 4bc - 4ac + 2c - 4ab + 2b + 2a - 1}{27abc - 9bc - 9ac + 3c - 9ab + 3b + 3a - 1}\right)$

$= 7^3\left(\dfrac{8abc - 4 (ab + ac + bc) + 2 (a + b + c) - 1}{27abc - 9 (ab + ac + bc) + 3 (a + b + c) - 1}\right)$

$= 7^3\left(\dfrac{-8 + 8 + 2 - 1}{-27 + 18 + 3 - 1}\right)$

$= 7^3\left(\dfrac{1}{-7}\right)$

$= -7^2 = \boxed{-49}$

Notes: This solution was obtained from the solutions of the linked question above. I didn't come up with this solution, I just took someone else's solution and applied it here. All credits go to the people who shared their solutions in the linked question.

I also have another solution in the linked question, where I fully expanded the product into an expression of 27 terms and substituted the values in. Not the best method available, but if you want to try out that method, you can refer to my solution in the linked question for details.