# Indeterminate is Algebraic over Transcendental Extension

I'm relatively new to the study of fields, and was presented with the following problem:

Let $$u = \frac{t^3}{t+1} \in K(t)$$. Show $$K(t)$$ is algebraic over $$K(u)$$ and determine $$[K(t):K(u)]$$.

I believe that this problem boils down to showing that $$t$$ is algebraic over $$K(u)$$ and that $$[K(t):K(u)]$$ follows from the degree of the minimal polynomial of $$t$$ over $$K(u)$$, but I am unsure of how to go about attacking this problem. I doubt that explicitly searching for a polynomial for which $$t$$ is a root is a viable strategy, especially when considering the case when $$u$$ is an arbitrary element of $$K(t)$$. I was told that using the unique factorization of polynomials will help, but I'm not immediately seeing why this would be useful.

What are some approaches to solving such a problem?

For the first part, rearrange the equation holding between $$t$$ and $$u$$ and thereby find out that $$t$$ is a root of the polynomial $$X^3-uX-u\in K(u)[X].$$ It follows that $$[K(t):K(u)]$$ is a divisor of $$3$$, so either $$3$$ or $$1$$. If the degree is $$1$$, we have $$K(t)=K(u)$$ and so $$t$$ must be a rational expression in $$u$$, $$t=\frac{f(u)}{g(u)}$$ with $$f,g\in K[X]$$ coprime and $$g\not\equiv 0$$. But then $$u=\frac{f(u)^3/g(u)^3}{1+f(u)/g(u)},$$ or $$f^3(u)=u g^2(u)(g(u)+f(u)).$$ As $$u$$ is transcendental over $$K$$, it follows that $$f^3(X)=X g^2(X)(g(X)+f(X)).$$ From the factor $$X$$ on the right, we conclude that $$X\mid f$$, hence $$X^2\mid g^2(X)(g(X)+f(X))$$. As $$g$$ is coprime to $$f$$, we have $$X\nmid g$$ and also $$X\nmid f+g$$, contradiction.