Find $\frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3}$ for $ax^3 + bx^2 + cx + d$

Question

Using Vieta's formulas, I can get $$\begin{align} \frac{1}{x_1^3} + \frac{1}{x_2^3} + \frac{1}{x_3^3} &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\&= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{x_1^3x_2^3x_3^3} \\ &= \frac{x_1^3x_2^3 + x_1^3x_3^3 + x_2^3x_3^3}{\left (-\frac{d}{a} \right)^3}\end{align}$$ But then I don't know how to substitute the numerator.

Answer 1

Let $y = 1/x$. So you ask for the value of $f(y) = y_1^3 + y_2^3 + y_3^3$ where the $y_i$ are solutions of $dy^3 + cy^2 + by + a = 0$. With this equation, write $$- d f(y) = c (y_1^2 + y_2^2 + y_3^2) + b (y_1 + y_2 + y_3) + 3 a.$$

Note the identity $$y_1^2 + y_2^2 + y_3^2=(y_1+y_2+y_3)^2-2(y_1y_2+y_1y_3+y_2y_3).$$ So you have

$$- d f(y) = c (\color{red}{y_1+y_2+y_3})^2-2c (\color{green}{y_1y_2+y_1y_3+y_2y_3}) + b (\color{red}{y_1 + y_2 + y_3}) + 3 a.$$

Now Vieta's formulae give (note the order of the constants!)

$$y_1+y_2+y_3 = -\frac cd\quad \text{and}\quad y_1y_2+y_1y_3+y_2y_3 = \frac bd.$$ So you obtain

$$- d f(y) = c \left(\color{red}{-\frac cd}\right)^2-2c \left(\color{green}{\frac bd}\right) + b \left(\color{red}{-\frac cd}\right) + 3 a,$$

hence the final result

$$ f(y) = y_1^3 + y_2^3 + y_3^3 = - \frac{c^3}{d^3} + \frac{3bc}{d^2} - \frac{3a}d = \frac{-c^3+ 3 bcd - 3 ad^2}{d^3}. $$

Answer 2

If $x \neq 0$ is a solution to $at^3+bt^2+ct+d=0$ then since $a+b(\frac{1}{x})+c(\frac{1}{x})^2+d(\frac{1}{x})^3=0$, $\frac{1}{x}$ is a solution to $dt^3+ct^2+bt+a=0$. Thus if we have $\frac{1}{x_i}=y_i$ for $i=1,2,3$,

\begin{align*} \frac{1}{x_1^3}+\frac{1}{x_2^3}+\frac{1}{x_3^3}&=y_1^3+y_2^3+y_3^3\\ &=(y_1+y_2+y_3)\left[(y_1+y_2+y_3)^2-3(y_1y_2+y_2y_3+y_3y_1)\right]+3y_1y_2y_3\\ &=(-\frac{c}{d})\left[(-\frac{c}{d})^2-3\cdot\frac{b}{d}\right]-3\cdot\frac{a}{d}\\ &=\frac{-c^3+3bcd-3ad^2}{d^3}. \end{align*}