Calculate $\dfrac{1}{x_1^3} + \dfrac{1}{x_2^3}$, where $x_1$ and $x_2$ are the roots of the equation $2x^2 - 3ax -2 = 0$.
Given quadratic equation: $\;$ $2x^2 - 3ax - 2 = 0$
The roots of the given quadratic equation are $\;$ $x_1$ $\;$ and $\;$ $x_2$.
Sum of roots $= x_1 + x_2 = \dfrac{3a}{2}$ $\;\;\; \cdots \; (1a)$
Product of roots $= x_1 \cdot x_2 = \dfrac{-2}{2} = -1$ $\;\;\; \cdots \; (1b)$
Now,
$\begin{aligned}
\dfrac{1}{x_1^3} + \dfrac{1}{x_2^3} & = \dfrac{x_1^3 + x_2^3}{x_1^3 \cdot x_2^3} \\\\
& = \dfrac{\left(x_1 + x_2\right) \left[x_1^2 - x_1 \cdot x_2 + x_2^2\right]}{x_1^3 \cdot x_2^3} \\\\
& = \dfrac{\left(x_1 + x_2\right) \left[\left(x_1 + x_ 2\right)^2 - 3 x_1 \cdot x_2\right]}{\left(x_1 \cdot x_2\right)^3} \\\\
& = \dfrac{\dfrac{3a}{2} \left[\left(\dfrac{3a}{2}\right)^2 - 3 \times \left(-1\right)\right]}{\left(-1\right)^3} \;\;\; \left[\text{by equations (1a) and (1b)}\right] \\\\
& = \dfrac{-3a}{2} \left[\dfrac{9a^2}{4} + 3\right] \\\\
& = \dfrac{-27a^3}{8} - \dfrac{9a}{2}
\end{aligned}$