If $\alpha, \beta, \gamma$ are the roots of the cubic $\ ax^3+bx^2+cx+d$, show that $\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$.

Question

I know that $\alpha + \beta + \gamma = -\frac{b}{a}%$ and that $\alpha\beta\gamma=-\frac{d}{a}$, but I don't know how to go from there to $\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$.

Answer 1

The given equation is $ ax^3+bx^2+cx+d=0$.

This equation can be written as

$ x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a}=0$ (on the assumption that $a \neq 0$)

This equation is a cubic equation and hence it has three roots $ \alpha, \beta, \gamma$, say.

Thus,

$\begin{align} x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} &=(x-\alpha)(x-\beta)(x-\gamma) \\ &= x^3-(\alpha+\beta+\gamma)x^2+(\alpha \beta+\beta \gamma+\gamma \alpha)x-\alpha \beta \gamma \end{align}$

Comparing the coefficient of $\ x \ $ from both sides , we get

$ \alpha \beta+\beta \gamma+\gamma \alpha=\frac{c}{a}$.

This proves your result.

Answer 2

Hint:

Notice that we have $$ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)$$

Answer 3

If the roots are $\alpha, \beta, \gamma$, then

$$(x-\alpha)(x-\beta)(x-\gamma)=0$$

Expanding the left hand side gets

$$x^3 -(\alpha + \beta + \gamma)x^2 + (\alpha\beta +\beta\gamma +\gamma\beta)x - \alpha\beta\gamma = 0$$

Compare coefficients of $x$ between this equation and the original cubic to see that

$$\alpha\beta + \beta\gamma + \gamma\alpha = c/a$$