Finding the Roots of a Polynomial with Vieta's Formulas

Question

$\alpha \ and\ \beta$ are two roots of $x^3+ax^2+bx+c=0$ Prove that $\alpha\beta$ is a root of $x^3-bx^2+acx-c^2=0$

Do you use Vieta's formulas?

Answer 1

Let $\alpha,\beta,\gamma$ be the roots of $x^3+ax^2+bx+c$. My claim is that $$ x^3-bx^2+acx-c^2 = (x-\alpha\beta)(x-\alpha\gamma)(x-\beta\gamma) \tag{1}$$ that boils down to proving that $b=\alpha\beta+\alpha\gamma+\beta\gamma$ (trivial by Vieta's formulas), that $c^2=\alpha^2\beta^2\gamma^2$ (again, trivial by Vieta's formulas) and that $$ ac = \alpha^2\beta\gamma+ \alpha\beta^2\gamma+\alpha\beta\gamma^2 \tag{2}$$ trivial again, since the RHS of $(2)$ is $(\alpha+\beta+\gamma)\alpha\beta\gamma = ac$.