Let $\displaystyle f(x)=x^3+x-1$ and $\displaystyle \theta$ be its real root
Express $\displaystyle \frac{\theta^7+4}{\theta+1}$ as a linear combination of $\displaystyle 1,\theta$ and $\displaystyle \theta^2$
note that:
$\displaystyle \theta^3 = 1 - \theta$.
hence:
$\displaystyle (\theta + 1)^3 = \theta^3 + 3\theta^2 + 3\theta + 1 = 1 - \theta + 3\theta^2 + 3\theta + 2 = 3\theta^2 + 2\theta + 2$
$\displaystyle = 3\theta^2 + 6\theta + 3 - 4\theta - 1 = 3(\theta + 1)^2 - 4(\theta + 1) + 3$
that is:
$\displaystyle 1 = \frac{1}{3}(4(\theta + 1) - 3(\theta + 1)^2 + (\theta + 1)^3)$
so that evidently:
$\displaystyle \frac{1}{\theta + 1} = \frac{1}{3}(4 - 3(\theta + 1) + (\theta + 1)^2) = \frac{1}{3}(2 - \theta + \theta^2)$
this is probably the hard part.
now:
$\displaystyle \theta^7 + 4 = (\theta^3)^2\theta + 4 = (1 - \theta)^2\theta + 4 = \theta - 2\theta^2 + \theta^3 + 4$
$\displaystyle =\theta - 2\theta^2 + 1 - \theta + 4 = 5 - 2\theta^2$
thus:
$\displaystyle \frac{\theta^7 +4}{\theta + 1} = \frac{1}{3}(5 - 2\theta^2)(2 - \theta + \theta^2)$
$\displaystyle = \frac{1}{3}(10 - 5\theta + \theta^2 + 2\theta^3 - 2\theta^4) = \frac{1}{3}(10 - 5\theta + \theta^2 + 2 - 2\theta - 2(1 - \theta)\theta)$
$\displaystyle = \frac{1}{3}(12 - 7\theta + \theta^2 - 2\theta + 2\theta^2) = \frac{1}{3}(12 - 9\theta + 3\theta^2) = 4 - 3\theta + \theta^2$
...if...my arithmetic is ok....
the trouble with this is you still have some remainder $\displaystyle \frac{r}{\theta + 1}$ so finding the inverse of $\displaystyle \frac{1}{\theta + 1}$ appears to be unavoidable.
Yep. Your arithmetic is ok.
Alternatively:$\displaystyle \theta^7 +4 = {(\theta^3)^2\theta +4} = {(1 - \theta)^2\theta +4} = {\theta^3-2\theta^2+\theta +4}$
Now do a long division by $\displaystyle \theta + 1$ to find (remainder turns out to be zero):$\displaystyle {\theta^7 +4 \over \theta + 1} = {\theta^3-2\theta^2+\theta +4 \over \theta + 1} = \theta^2-3\theta+4. \quad \Box$