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.

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