$K=F_3(t)$ show that for every root $\alpha$ of the polynomial $f(x)=x^6-tx^3+1$, $K(\alpha)/K$ is a splliting field

Question

$K=F_3(t)$ show that for every root $\alpha$ of the polynomial $f(x)=x^6-tx^3+1 \in K[x] $ , $K(\alpha)/K$ is a splitting field.

By putting $u=x^3$ I showed that we have two roots (because either $t=\pm 2$ but $\pm 2 \in F_3 $ so $t$ cannot extend $F_3$) for every $a\in F_3$ I know that $x^3=x$, but this condition doesn't refer tor $t$.

So how could I show that this extension is splitting with one of the roots $\alpha$ ?

Answer 1

Observe that $$\begin{align}f(x)&=x^6+2tx^3+1\\&=(x^3+t)^2+1-t^2\\&=(x^3+t-\sqrt{t^2-1})(x^3+t+\sqrt{t^2-1})\\&=(x+\sqrt[3]{t-\sqrt{t^2-1}})^3(x+\sqrt[3]{t+\sqrt{t^2-1}})^3\end{align}$$ This means that $f$ has only two roots $\alpha,\beta$ and they satisfy $(\alpha+\beta)^3=\alpha^3+\beta^3=2t$.

If $f(\alpha)=0$, then $2t=\frac{\alpha^6+1}{-\alpha^3}=((\alpha^2+1)/2\alpha)^3$. This means that $\sqrt[3]{2t}\in K(\alpha)$.

But then $\beta=\sqrt[3]{2t}-\alpha\in K(\alpha)$.