Cyclic summation of roots of a cubic

Note that $\alpha\beta+\beta\gamma+\gamma\alpha = 0$

$\large \left | \displaystyle \sum_{\text{cyc}} \left( \dfrac{\alpha}{\beta} + \dfrac{\beta}{\alpha} \right )\right| = \large \left| \frac{{\alpha}^2\beta+{\alpha}^2\gamma+{\beta}^2\alpha+{\beta}^2\gamma+{\gamma}^2\alpha+{\gamma}^2\beta}{\alpha\beta\gamma} \right|$

$= \large \left| \frac{(\alpha\beta+\beta\gamma+\gamma\alpha)(\alpha+\beta+\gamma)-3\alpha\beta\gamma}{\alpha\beta\gamma} \right|$

$= \large \left| \frac{-3\alpha\beta\gamma}{\alpha\beta\gamma} \right|$

$= 3$

Given that $\alpha = a, \beta=b, \gamma=c$

Using Vieta's formulas you obtain these three equations:

$abc=7/2$

$ab+bc+ac=0$

$a+b+c=-1/2$

From the first equation we obtain that:

$ab=7/2c$

$bc=7/2a$

$ac=7/2b$

This means that:

$1/a+1/b+1/c=0$

The solution is given by:

$(1/a+1/b+1/c)*(a+b+c)=0$

$a/b+b/a+a/c+c/a+b/c+c/b+a/a+b/b+c/c=0$

$(a/b+b/a)+(a/c+c/a)+(b/c+c/b)=-3$

$\mid-3\mid=\boxed{3}$