Show a rational function is transcendental over a field. 
Let $u=\frac{x^3}{x+1}\in F(x)$, where $F(x)$ is the field of quotients of $F[x]$ ($F$ some field, $x$ an indeterminate over it). Show that $u$ is transcendental over $F$.

This is an exercise in Hungerford.
I'm having some trouble even grasping the concepts involved.  For instance, I know that if $v$ is transc. over $F$, then $F[v]\cong F[x]$. Or that if $v$ is transc. over $F$, then $F[v]\subsetneq F(v)$.  But I have no idea how to use this to my advantage.  
I'm also confused about what it even means for $u$ as above to be transc. over $F$.  Am I going to have to consider "polynomials of polynomials"?
 A: Since this is an important basic issue, I'll add a complementary answer to Bill Dubuque's, and the good comments above: the definition of $u$ shows that $x$ satisfies a cubic equation over $F(u)$, so is algebraic over $F(u)$. If $u$ were algebraic over $F$, then, by transitivity of "algebraic extension", $x$ itself would be algebraic over $F$.
This less-explicit but more-qualitative kind of argument can succeed when explicit computations become burdensome.
A: The following way of saying it is a bit more concrete, in case anyone prefers it that way.
If the coefficients $c_k$ in the sum
$$
\sum_{k=0}^n c_k u^k
$$
are in $F$, then the point is to show that that sum cannot be $0$ unless all of the coefficients are $0$.  The sum is
$$
\sum_{k=0}^n c_k \left( \frac{x^3}{x+1} \right)^k.
$$
If that $=0$ then multiplying both sides by $(x+1)^n$, one concludes
$$
\sum_{k=0}^n c_k x^{3k} (x+1)^{n-k} = 0.
$$
So a polynomial in $x$ evaluates to $0$.  Since $x$ is transcendental, that can't happen unless all of the coefficients are $0$.
