Show that this extension is finite?

Question

I have to show that $ k (\frac {x^3+2}{x^2+x-1} ) \subset k(x)$ is an algebraic extension and have to find a basis.

$(x^2+x-1)$ does not divide ($x^3+2$) perfectly.
$k(x)$ is a vector space over k $ \frac {x^3+2}{x^2+x-1}$.

I can't see any way ahead.
Can someone suggest how to get started on this. Any suggestion are welcome.

Answer 1

Let $(x^3+2)/(x^2+x-1) = a$. Then the polynomial $a(y^2+y-1)-y^3-2$ is a polynomial with coefficients in $ k(a)$ s.t. $x $ is a root of this polynomial. Hence it is algebraic. Now can you find basis?

Answer 2

The only thing you have to do is to find a polynomial with coefficients in the field $k(\frac{x^3+2}{x^2+x-1})$ such that $x$ is a root of this polynomial.

Answer 3

This is a kind of question that feels either completely mysterious or completely obvious, with no level in-between, so it is a bit hard to give directions.

If you were given a number $y \in \Bbb C$, how would you go about solving for $x$ in $\frac {x^3+2}{x^2+x-1} = y$ ?
How many solutions should there be in $\Bbb C$ ?
Is there a point in your work where you have written down a polynomial equation in $x$ and $y$ ?