Question:
Compute the Galois group of the polynomial $x^3+t^2x-t^3$ over $\mathbb{C}(t)$ where $\mathbb{C}(t)$ is the field of rational functions in one variable over complex numbers $\mathbb{C}$.
Answer:
Let $f(x)=x^3+t^2x-t^3$. Then suppose $f(kt)=(kt)^3+t^2(kt)-t^3=0$. Thus, we obtain $k^3t^3+kt^3-t^3=0$, which yields that $k^3+k-1=0$. Now in $\mathbb{C}$, I can find three $k$ (not necessarily distinct) satisfying that relation. This shows that in fact $f$ splits into linear factors in $\mathbb{C}(t)$. Hence, the splitting field of $f$ over $\mathbb{C}(t)$ must be $\mathbb{C}(t)$ itself. Therefore, $Gal(f)$ must be the trivial group.
Do you see any mistake in my proof? Thanks in advance...
Looks good to me.
Because $f$ is homogeneous of degree three as a polynomial in two variables, it factors as $$f(x)=(x-k_1t)(x-k_2t)(x-k_3t)$$ in the ring $\Bbb{C}(t)[x]$. Here $k_1,k_2,k_3$ are the complex zeros of the equation $$k^3-k+1=0$$ that you used. The roots $k_it$, $i=1,2,3$, are all in the base field $\Bbb{C}(t)$, so $\Bbb{C}(t)$ itself is the splitting field and the Galois group is trivial.