Question: In an exam there is given the general equation for quadratic: $ax^2+bx+c=0$.

It is asking: what does $\dfrac{1}{{x\_1}^3}+\dfrac{1}{{x\_2}^3}$ equal?

Answer: 
BIG HINT:
$$\frac{1}{x\_1^3}+\frac{1}{x\_2^3}=\frac{x\_1^3+x\_2^3}{x\_1^3x\_2^3}=\frac{(x\_1+x\_2)(x\_1^2-x\_1x\_2+x\_2^2)}{x\_1^3x\_2^3}=\frac{(x\_1+x\_2)((x\_1+x\_2)^2-3x\_1x\_2)}{(x\_1x\_2)^3}$$
If $x\_{1}$ and $x\_{2}$ are the roots then

$x\_{1} + x\_{2} = -\frac{b}{a}$ and $x\_{1} \cdot x\_{2}=\frac{c}{a}$, now $\frac{1}{x\_{1}}+\frac{1}{x\_{2}} = -\frac{b}{c}$ and $$\frac{1}{x\_{1}^{3}}+\frac{1}{x\_{2}^{3}} = \left(\frac{1}{x\_{1}}+\frac{1}{x\_{2}}\right)^3- 3\cdot\frac{1}{x\_{1}.x\_{2}}\left(\frac{1}{x\_{1}}+\frac{1}{x\_{2}}\right) = \frac{3abc-b^{3}}{c^{3}}$$