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: 
Let $\dfrac1{x^3}=y\iff x^3=?$

$$(-c)^3=(ax^2+bx)^3\iff -c^3=a^3(x^3)^2+b^3(x^3)+3ab(x^3)(-c)$$

$$\iff-c^3=\dfrac{a^3}{y^2}+\dfrac{b^3}y-\dfrac{3abc}y$$

$$\iff c^3y^2+(b^3-3abc)y+a^3=0$$ whose roots are $\dfrac1{x\_1^3},\dfrac1{x\_2^3}$

Can you apply Vieta's formula now?
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 $$\frac{1}{x\_{1}^{3}}+\frac{1}{x\_{2}^{3}} = \left(\frac{1}{x\_{1}}+\frac{1}{x\_{2}}\right)^3- 3\cdot\frac{1}{x\_{1}.x\_{2}}\left(\frac{1}{x\_{1}}+\frac{1}{x\_{2}}\right) = \frac{3abc-b^{3}}{c^{3}}$$