I need to show that
$t^{6}-2t^{3}-1$ is irreducible over $\mathbb{Q}$.
I tried to express it, in three different ways, such:
$t^{6}-2t^{3}-1= (t^{3}+a)(t^{3}+b)$. In this way, I got a contradiction.
$t^{6}-2t^{3}-1= (t^{4}+at^{3}+bt^{2}+ct +d)(t^{2}+\alpha t + \beta)$. In this way, there are many combinations.
$t^{6}-2t^{3}-1= (t^{5}+at^{4}+bt^{3}+ct^{2} +dt +e)(t+\alpha )$. I think, I will get a contradiction.
There is an easier way to show that it is irreducible ? Thanks
Pd. I also tried the eisenstein's criterion, but it can't apply for this.
My immediate thought is to show that $$\mathbb{Q}[t]/(t^6 - 2t^3 - 1)$$ is a domain. This is tough to do directly, but if we first look at a field extension $K / \mathbb{Q}$ where $K = \mathbb{Q}[x] / (x^2 - 2x - 1)$ which is absolutely a field (since $x^2 - 2x- 1$ is irreducible). Then what is $t$? It's just a third root of $x$, so $$\mathbb{Q}[t]/(t^6 - 2t^3 -1) = K[t]/(x-t^3)$$ And since $K$ is a field and $x-t^3$ is irreducible, $\mathbb{Q}[t]/(t^6-2t^3 - 1)$ is a field.
Obviously you'll have to check some details.