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 $-\dfrac{11}{3}$ 
 A: $$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}$$
$$= \frac{\alpha + \beta + \omega - \omega}{\omega} + \frac{\beta + \omega + \alpha - \alpha}{\alpha} + \frac{\alpha + \omega + \beta - \beta}{\beta}$$
$$ = (\alpha + \beta + \omega) \left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\omega}\right) - 3$$
$$ = (\alpha + \beta + \omega) \left(\frac{\beta\omega}{\alpha\beta\omega} + \frac{\alpha\omega}{\alpha\beta\omega} + \frac{\alpha\beta}{\alpha\beta\omega}\right) - 3$$
$$ = \frac{\alpha + \beta + \omega}{\alpha\beta\omega}(\beta\omega + \alpha\omega + \alpha\beta) - 3$$
I think you should be able to take it from there.
A: Alternatively, you can solve the equation:
$$x^3-4x^2+x+6=0 \Rightarrow (x+1)(x-2)(x-3)=0 \Rightarrow \\
\alpha =-1, \beta =2,\omega=3.$$
Hence: 
$$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}=\\
\frac{-1+ 2}{3} + \frac{2 + 3}{-1} + \frac{-1 + 3}{2}=\\
\frac13-5+1=\\
-\frac{11}{3}.$$
A: Hint: We can write $$\frac{4-w}{w}+\frac{4-\beta}{\beta}+\frac{4-\alpha}{\alpha}$$ and this is $$4\left(\frac{\alpha\beta+\alpha w+w\beta}{\alpha \beta w}\right)-3$$ and this is $$-\frac{2}{3}\left(1-\beta w-\alpha w+\alpha w+\beta w\right)$$
This simplifies to $$-\frac{2}{3}-3=-\frac{11}{3}$$
A: That follows from your results,  since we get:  $\dfrac{4-\omega}{\omega}+\dfrac{4-\beta}{\beta}+\dfrac{4-\alpha}{\alpha}=\dfrac{4(\omega\beta+\omega \alpha+\beta\alpha)-3\omega\beta\alpha}{\omega \beta \alpha}=\dfrac{4+18}{-6}=-\dfrac{11}3$.
