<s>[INST] Solving polynominals equations (relationship of roots) 
The roots of $x^3-4x^2+x+6$ are $\alpha$, $\beta$, and $\omega$.
  Find (evaluate): 
  $$\frac{\alpha+\beta}{\omega}+\frac{\alpha+\omega}{\beta}+\frac{\beta+\omega}{\alpha}$$

So far I have found:
$$\alpha+\beta+\omega=\frac{-b}{a} = 4 \\
\alpha\beta+\beta\omega+\alpha\omega=\frac{c}{a} = 1 \\
\alpha×\beta×\omega=\frac{-d}{a} = -6$$
And evaluated the above fractions creating
$$\frac{\alpha^2\beta+\alpha\beta^2+\alpha^2\omega+\alpha\omega^2+\beta^2\omega+\beta\omega^2}{\alpha\beta\omega}$$ 
I don't know how to continue evaluating the question.
Note:
The answer I have been given is $-\dfrac{11}{3}$ 
 [/INST] $$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}$$
$$= \frac{\alpha + \beta + \omega - \omega}{\omega} + \frac{\beta + \omega + \alpha - \alpha}{\alpha} + \frac{\alpha + \omega + \beta - \beta}{\beta}$$
$$ = (\alpha + \beta + \omega) \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\omega}\right) - 3$$
$$ = (\alpha + \beta + \omega) \left(\frac{\beta\omega}{\alpha\beta\omega} + \frac{\alpha\omega}{\alpha\beta\omega} + \frac{\alpha\beta}{\alpha\beta\omega}\right) - 3$$
$$ = \frac{\alpha + \beta + \omega}{\alpha\beta\omega}(\beta\omega + \alpha\omega + \alpha\beta) - 3$$
I think you should be able to take it from there.
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