Possible pattern involving $x$ in the continued fraction expansion of $\frac{1}{\sqrt[3]{x^3+1}-x}$ Consider the expression $$\frac{1}{\sqrt[3]{x^3+1}-x}$$
Plugging in $10$ for $x$ and using $W|A$, we find that the continued fraction expansion is $[300; \mathbf{10}, 450, 8, ...]$.
For $x=11$, it's $[363; \mathbf{11}, 544, ...]$. The pattern continues (I have only checked up to $21$, but I assume that it's true) that the second term in the continued fraction expansion is $x$.
I have two questions:

*

*Why does this pattern occur and how do we prove that it occurs?

*Is there a similar pattern with later terms?

I haven't found an answer to 2, and I've had no luck as of yet for 1. All I have done so far is simplify it to $x\sqrt[3]{x^3+1}+\sqrt[3]{(x^3+1)^2}+x^2$.
Thanks in advance.

Let $$\sqrt[3]{x^3+1} - x = a$$  Then  $$a^3 + 3a^2x + 3ax^2 = 1$$ This shows that  $a < \frac{1}{3x^2}$.  Now we have everything we need to work out the continued fraction:
$$\begin{align*} \frac{1}{a} &= 3x^2 + a(3x + a)\\
&=3x^2 + \frac{1}{\frac{1}{a}\frac{1}{3x+a}}\\
&=3x^2 + \frac{1}{\frac{3x^2 + 3ax + a^2}{3x+a}}\\
&=3x^2 + \frac{1}{x + a\frac{2x + a}{3x+a}}\\
 \end{align*}$$
Using the upper bound $a<\frac{1}{3x^2}$, you can show that $a\frac{2x + a}{3x+a} < 1$, which proves the relationship you observed.
You can continue this process to find more continued fraction terms.  $$a\frac{2x + a}{3x+a} = \frac{1}{\frac{1}{a}\frac{3x+a}{2x+a}}$$
Again, replace $\frac{1}{a} = 3x^2 + 3ax + a^2$ and separate into term independent of $a$ and term proportionate to $a$
