If $\alpha$, $\beta$, $\gamma$ are the zeros of $x^3 + 4x + 1=0$, compute a expression containing them If $\alpha$, $\beta$, $\gamma$ are the zeros of $x^3 + 4x + 1$, then calculate the value of: $$(\alpha + \beta)–1 + (\beta + \gamma)–1 + (\gamma + \alpha)–1$$
 A: HINT: You want to calculates $2(\alpha+\beta+\gamma)-3$. There is a formula to find $\alpha+\beta+\gamma$ when you know the coeficients of the polynomial.
A: Note that $(\alpha + \beta)-1+(\beta+\gamma)-1+(\gamma +\alpha)-1=\left(2\alpha+2\beta+2\gamma\right)-3=2(\alpha+\beta+\gamma)-3$. We know from Vieta's formulae that: 
$$\alpha+\beta+\gamma=-\frac{0}{1}=0$$
Therefore, $(α + β)–1 + (β + γ)–1 + (γ + α)–1=2(0)-3=-3$.
A: Hint:
Sum of the roots of equation : $x^3+bx^2+cx+d=0$ is $-b$.
So convert it into this form. You see that co-efficient of $x^2$ is zero, so the expression's value is $-3$.
Hail Vieta!!
A: Hint $\ $ Exploit symmetry. The sum $\rm = 2(a+b+c) - 3,\:$ and $\rm\:a+b+c = $ -coef of $\rm\,x^2\,$ by Vieta.
Remark $\ $ Generally one can calculate any symmetric polynomial of the roots in the same way, since, by a simple algorithm of Gauss, any symmetric polynomial may be written as a polynomial in the elementary symmetric polynomials  (Fundamental Theorem of Symmetric Polynomials) 
A: Hint: So you want $2(\alpha + \beta + \gamma) - 3$ ?
