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 $-\dfrac{11}{3}$ 

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: 
$$\frac{\alpha + \beta}{\omega} + \frac{\beta + \omega}{\alpha} + \frac{\alpha + \omega}{\beta}=\\
\frac{-1+ 2}{3} + \frac{2 + 3}{-1} + \frac{-1 + 3}{2}=\\
\frac13-5+1=\\
-\frac{11}{3}.$$
