Prove $\alpha^3 + \beta^3 = S^3 - 3PS$ in Quadratic Equation

Question

We have a quadratic equation like this: $ax^2 + bx + c = 0$ and we know that $S = \alpha + \beta = -b/a$ and $P = \alpha\beta = c/a$. How we can prove that $\alpha^3 + \beta^3 = S^3 - 3PS$ and is there any relation to $\alpha^n + \beta^n$?

Answer 1

Hint: expand $(\alpha + \beta)^3$ and factor everything except for $\alpha^3$ and $\beta^3$.

Answer 2

We have $$ax^2+bx=-c\implies (-c)^3=(ax^2+bx)^3$$

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

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

So, $\displaystyle\alpha^3,\beta^3$ are the roots of $$a^3y^2+y(b^3-3abc)+c^3=0$$