Quadratic equation, find $1/x_1^3+1/x_2^3$

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 1

BIG HINT: $$\frac{1}{x_1^3}+\frac{1}{x_2^3}=\frac{x_1^3+x_2^3}{x_1^3x_2^3}=\frac{(x_1+x_2)(x_1^2-x_1x_2+x_2^2)}{x_1^3x_2^3}=\frac{(x_1+x_2)((x_1+x_2)^2-3x_1x_2)}{(x_1x_2)^3}$$

Answer 2

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 $$\frac{1}{x_{1}^{3}}+\frac{1}{x_{2}^{3}} = \left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right)^3- 3\cdot\frac{1}{x_{1}.x_{2}}\left(\frac{1}{x_{1}}+\frac{1}{x_{2}}\right) = \frac{3abc-b^{3}}{c^{3}}$$

Answer 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?