<s>[INST] Degree of splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$ I'm trying to find the degree splitting field of $x^6+tx^3+t$ in $\mathbb{F}_3(t)[x]$. After substituting $z= x^3$, and using the qudratic formula, and thensubstituting $x$ back in, I get the roots
$$\frac{\sqrt[3]{-t\pm\sqrt{t^2-4 t}}}{\sqrt[3]{2}},$$
but how do I find the minimal polynomial for this? No matter what power I take, I can't seem to get rid of the radicals. Is there another way to think about square/cube roots when you are in a finite field? Not sure how to proceed.
 [/INST] As @jyre says, the polynomial $X^6 + tX^3 + t$ is irreducible, so you can make a first field extension of degree $6$. $\Bbb F_3(t) \to k = \Bbb F_3(t)[x]/(x^6+tx^3+t)$.
Over this field $k$, $x$ is a root of $X^6 + tX^3 + t$ so the polynomial factors a bit. It turns out $x$ is a multiple root :  
$X^6 + tX^3 + t = X^6 + tX^3 +t - (x^6 + tx^3 + t) 
\\= X^6-x^6 + t(X^3-x^3) = (X^3-x^3)(X^3+x^3+t) 
\\= (X-x)^3(X^3+x^3+t)$
Now you need to check if $X^3+x^3+t$ is further reducible over $k$.
But putting $t = -x^6/(x^3+1)$, $k$ is isomorphic to $\Bbb F_3(x)$, and now the remaining cubic factor is $X^3 + x^3-x^6/(x^3+1) = X^3+x^3/(x^3+1) = (X+x/(x+1))^3$
Therefore, your polynomial splits in $k$ as $(X-x)^3(X+x/(x+1))^3$
And so the splitting field is $k$, and the degree is $6$.
</s>