Let $x$ be an indeterminate over $\mathbb{C}$ and $y=\frac{x^3}{x+1}$. Find the minimal polynomial $f(z)$ of $x$ over $\mathbb{C}(y)$. What is the Galois group of $f(z)$ over $\mathbb{C}(y)$
I was able to find the minimal polynomial but stuck in proving that its actually minimal. Apart from that I have no idea how to compute the Galois Group.
Computations of Minimal Polynomial
Suppose $z=x$, then $\frac{z^3}{z+1}=\frac{x^3}{x+1}=y$ and hence I feel that the minimal polynomial is $z^3-yz-y$. Clearly this polynomial satisfied by $x$ and all we want to do is prove its minimality.
If it is no minimal, then it is divisible by the minimal polynomial of $x$, so $z^3-yz-y$ is reducible and hence it has a linear factor and hence some element $g(y) \in \mathbb{C}(y)$ satisfies this polynomial. Suppose $g(y)=a_ny^n+a_{n-1}y^{n-1} \dots a_1y+a_0$ is the root of this polynomial. Then
$$(g(y))^3-yg(y)-y=0$$ which inturn implies
$$(g(y))^3=y\big(g(y)+1\big)$$ Multiplying both sides by $(x+1)^{3n}$ gives $$\Big(\sum_{i=0}^{n}a_i(x^3)^i(x+1)^{n-i} \Big)^3=(x+1)^{3n-2}(x^3)(x^3-x-1)$$
Comparision of the degree gives us that $6i+3n=3n+4$ which is impossible because $i$ takes integer values. So I reached a contradiction and hence $f(z)$ is irreducible.
Computation of Galois Group.
I haven't been able to proceed much. All I know is
$$f(z)=(z-x)(z^2+zx+x^2-y)$$ and using quadratic formula shows that the $\sqrt{x^2-4x^2+4y} $ must be there in the splitting field.
So request someone to verify the proof of minimal polynomial and also to guide me for the computation of Galois Group. Thanks.