Let $F$ be a field and let $\text{Frac}( F[x])$ be the field of rational polynomials over $F$. Let $a = \frac{x^3}{x+1}$. Show that $\text{Frac}( F[x])$ is an algebraic extension over $F(a)$.
I have this so far:
$x$ is algebraic over $F(a)$ since $x^3-ax-a = 0$, and thus every other rational polynomial is algebraic over $F(a,x)$ since if $\frac{F(x)}{G(x)} \in \text{Frac}( F[x])$ then $$ \frac{F(x)}{G(x)} * G(x) - F(x) = 0 $$ In simple words, if we can show $x$ is algebraic starting from $a$, then we can construct any polynomial starting from $x$ ,and then every rational expression would be solution to a polynomial over $F(a,x)$ Then $[\text{Frac} F[x]:F(a,x)] = 1$ and we get that $$[\text{Frac} F[x]:F(a)] = [F(a,x):F(a)].$$ Finally, I claim that this equals $3$ because $x^3-ax-a$ is the minimal polynomial of $x$ over $F(a)$, but I'm not able to show that it is irreducible. Can anybody help me?
Shorter version: Is this polynomial irrreducible over $F(a)$? $$f(t) = t^3 -at -a$$ where $a = \frac{x^3}{x+1}$.