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:
$$\frac{1}{x\_1^3}+\frac{1}{x\_2^3}=\frac{x\_1^3+x\_2^3}{x\_1^3x\_2^3}=\frac{(x\_1+x\_2)(x\_1^2-x\_1x\_2+x\_2^2)}{x\_1^3x\_2^3}=\frac{(x\_1+x\_2)((x\_1+x\_2)^2-3x\_1x\_2)}{(x\_1x\_2)^3}$$
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?