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

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*}
