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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?

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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.

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