I need to prove that $x^5-2x^3+2x-2$ is irreducible over $\mathbb{Q}(\sqrt[3]{2})$.
I know that it is irreducible over $\mathbb{Q}$. Suppose $\alpha\in\mathbb{Q}(\sqrt[3]{2})$, then:
$[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]$ = 3 = $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(\alpha)][\mathbb{Q}(\alpha):\mathbb{Q}]$ which implies that $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(\alpha)]=1$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 3$ or $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}(\alpha)]=3$ and $[\mathbb{Q}(\alpha):\mathbb{Q}] = 1$. But if we have $[\mathbb{Q}(\alpha):\mathbb{Q}] = 1$ then $\mathbb{Q}(\alpha) = \mathbb{Q}$ which is not possible because $\alpha\notin \mathbb{Q}$. Then we have that $[\mathbb{Q}(\alpha):\mathbb{Q}]=3$$\implies$the degree of the minimal polynomial of $\alpha $ over $\mathbb{Q}$,$f_\alpha$ has degree 3. Since $f_\alpha|f$,we would have that $f= (x^3+b_1x^2+c_1x+d_1)(x^2+c_2x+d_2)$ and this is not possible because $f$ is irreducible over $\mathbb{Q}$.
Is this proof correct ?
It does look as if you’re showing that a root of $f(x)=x^5 -2x^3+2x-2$ can’t lie in $\Bbb Q(\sqrt[3]2\,)$. But as @Arthur has correctly observed, this does not show irreducibility.
I’ll give an argument involving something fairly advanced, even though an elementary argument on the style of the proof you know for the Eisenstein criterion could be cobbled up.
With respect to the (unique) $2$-adic valuation $v$ on $K=\Bbb Q(\sqrt[3]2\,)$ in which we give $v(\sqrt[3]2)=1$, the Newton Polygon of $f$ has only the two vertices $(0,3)$ and $(5,0)$. The unique segment joining them passes through no other integral points than those endpoints.
This is enough to show that $f$ is $K$-irreducible, but I’ll give another argument, still N-Polygon-theoretic. All roots of $f$ must have $v$-valuation $3/5$. Any linear monic polynomial with such a root will have constant term of valuation $3/5$. Impossible over $K$. Similarly, any quadratic monic polynomial with two such roots must have constant term of valuation $6/5$, equally impossible over $K$.