In the field $F(u)$, let $\alpha = \frac{u^3}{(u+1)}$. Consider the subfield $F(\alpha)$ of $F(u)$.
Prove that the polynomial $f(x)=x^3-\alpha x-\alpha$ is irreducible over the field $F(\alpha)$.
Edit: We are assuming this is not a trivial extension. That is, assume $u \notin F(\alpha)$.
We have shown $u$ is a root of $f(x)$. We have thought about assuming the polynomial is reducible as $(x-u)(a+bx+cx^2)$, finding the coefficients a, b, and c, and coming to a contradiction. The problem is that we cannot find the contradiction.
Here I'm assuming $u$ is transcendental over $F$.
The roots of $x^3 - \alpha x - \alpha$ can be computed (e.g. using Mathematica) as
$$u, \frac{-u(u + 1)-u\sqrt{(u + 1)(u-3)} }{2 (u+1)}, \frac{-u(u + 1)+u\sqrt{(u + 1)(u-3)} }{2 (u+1)}$$
and you want to argue that those last two roots are not in $F(u)$ (and therefore not in $F(\alpha)$).
Here is one way to do this and this is not the only way to show that your polynomial is irreducible (Eisenstein's criterion should work as well).
The point is to argue that those square roots can't exist inside $F(u)$ and we can do this using the notion of valuations. The specific valuation we want is the valuation at $u = -1$ (which we call $v_{-1} : F(u) \to \mathbf{Z}$) which is defined for a polynomial $f(u)$ as the multiplicity of $-1$ as a root of $f$. E.g.
$$ v_{-1}((u + 1)^r) = r, \quad v_{-1}(u) = 0,\quad v_{-1}(u(u+1)) = 1.$$
We define $v_{-1}$ for a rational function $f/g$ as $v_{-1}(f/g) = v_{-1}(f) - v_{-1}(g)$.
In particular,
$$ v_{-1}(\alpha) = v_{-1}(u^3) - v_{-1}(u + 1) = -1. $$
Now the fun bit is that the valuations of roots of a polynomial are entirely determined by its Newton polygon.
We construct the Newton polygon in two steps. First, we take each term of the polynomial, e.g. $a_n x^n$ and plot a point $(n, v_{-1}(a_n))$ in the xy-plane. Second, we imagine taking a string, tying it to the leftmost point and wrapping it around the bottom of the points:
For $x^3 - \alpha x - \alpha$, we have three points: $(0, -1)$, $(1,-1)$, and $(3,0)$ which we get since the constant term ($x^0$) has valuation $-1$, the linear term ($x^1$) has valuation $-1$ and the cubic term ($x^3$) has valuation $0$.
Now the rule is that if you take the negative slopes of the Newton polygon, those tell you what the valuations of the roots can be and the length of the corresponding edge when projected onto the $x$-axis tells you the multiplicity.
For $x^3 - \alpha x - \alpha$ and the valuation $v_{-1}$ we have two edges in the polygon: $(0,-1) \to (1,-1)$ and $(1,-1) \to (3,0)$. The first has slope $0$ and length $1$. The second has slope $\frac12$ and length $2$. This implies $x^3 - \alpha x - \alpha$ has 1 root with valuation $0$ at $u = 1$ and 2 roots with valuation $-\frac12$ at $u = 1$ (those are the two roots with the square roots).
Even without Newton polygons, it is not too hard to see that
$$ v_{-1}\left(\frac{-u(u + 1)\pm u\sqrt{(u + 1)(u-3)} }{2 (u+1)}\right) = -\frac{1}2 $$
and this is a problem since no polynomial can halve a root with multiplicity $\frac12$. This is something else you can show by hand: that
$$ \sqrt{u + 1} \notin F(u). $$
(This should be similar to showing that $\sqrt{2}$ is irrational.)
The advantage of using the Newton polygon is you don't need to solve for the other two roots to know they can't be in $F(u)$. The advantage of solving for the other roots and showing $\sqrt{u + 1} \notin F(u)$ is that you don't need to know about valuations or Newton polygons.