Use an iteration to approximate the specified cubic root

The root in $(0,1)$ is $\alpha\approx0.1674491911$, so $\alpha=0.16745$ to 5 d.p. Other roots of the cubic are not the root selected in this question.

The source calculator iterates the recurrence from $x_1=0$ and reports increasing convergence to $\alpha$. The reproduced values are:

The first two nonzero exact values are $x_2=1/6$, $x_3=217/1296$. Repeated rounded values are not a proof of exact equality or of convergence.

If $x_n\to\alpha$, then $x_{n+1}\to\alpha$ and $\alpha=(\alpha^3+1)/6$. Hence $\alpha^3-6\alpha+1=0$, as required. For this starting value, increasing bounded convergence can also be checked: the map is increasing on $[0,\alpha]$, sends that interval into itself, and $(x^3+1)/6>x$ below $\alpha$. The conditional limit calculation alone is not a proof for arbitrary starting values.