What's the easiest way to factor this cubic term?
Solution 1:
To combine the first two comments:
The polynomial is homogeneous in degree $3$ so if $x=a/b$ then you are trying to factor $$b^3(x^3+6x^2+11x+6)$$
For the bracketed expression, you might either
- spot that the odd and even coefficients sum to the same value with $1+11=6+6$ so $(x+1)$ is a factor, i.e. $(a+b)$ is a factor of the original expression, or
- hope there are rational factors, which by the rational root theorem would have to be of the form $(x\pm1),(x\pm2),(x\pm3),(x\pm6)$, and testing in this case shows three of these work