If $\alpha, \beta, \gamma$ are the roots of the cubic $\ ax^3+bx^2+cx+d$, show that $\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$. 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 $\alpha\beta+\alpha\gamma+\beta\gamma=\frac{c}{a}$.
 A: Hint:
Notice that we have $$ax^3+bx^2+cx+d=a(x-\alpha)(x-\beta)(x-\gamma)$$
A: 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
$$\alpha\beta + \beta\gamma + \gamma\alpha = c/a$$
A: 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. 
