$\mathbb{Q}(\sqrt[3]{17})$ has class number $1$

The first thing to do is get a clean basis without annoying congruence conditions : put $\beta=\frac{\alpha^2-\alpha+1}{3}$. Then $[1,\alpha,\beta]$ is a $\mathbb Z$-basis of $\mathcal{O}_K$.

We apply Franz Lemmermeyer's method and look for elements of the form $x+y\alpha+z\beta$ with interesting norms and $x,y,z$ small.

A little inspection shows that

$$ N(2+\alpha+\beta)=N(2-\beta)=2, N(1+\alpha+\beta)=N(1-\alpha+\beta)=3, N(3-\alpha)=2 \times 5 $$

A few additional checks and computations from here then reveals that

$$ \begin{array}{lclcl} \mathfrak{p}_1 &=& (2+\alpha+\beta) &=& \Bigg( \frac{\alpha^2+2\alpha+7}{3}\Bigg) \\ \mathfrak{p}_2 &=& (2-\beta) &=& \Bigg( \frac{-\alpha^2+\alpha+5}{3}\Bigg) \\ \mathfrak{p}_3 &=& (1+\alpha+\beta) &=& \Bigg( \frac{\alpha^2+2\alpha+4}{3}\Bigg) \\ \mathfrak{p}_4 &=& (1-\alpha+\beta) &=& \Bigg( \frac{\alpha^2-4\alpha+4}{3}\Bigg) \\ \mathfrak{p}_5 &=& (\frac{3-\alpha}{2+\alpha+\beta}) &=& \Bigg( \frac{-2\alpha^2-\alpha+16}{3}\Bigg) \\ \mathfrak{p}_6 &=& (\frac{5(2+\alpha+\beta)}{3-\alpha}) &=& \Bigg( \frac{11\alpha^2+28\alpha+74}{3}\Bigg) \\ \end{array} $$