<s>[INST] Cube root of two $\sqrt[3]2$ continued fraction I know there is a nice way of getting the continued fraction expansion of quadratic irrationals mainly because they recur after a point, and if they recur after a point they are quadratic irrationals. When constructing the expansion you can multiply by conjugates (kind of), e.g. 
$\sqrt 3 =1+\sqrt 3 -1 = 1+\frac {1}{\frac {\sqrt 3 +1}{2}} $
Where you use $(\sqrt 3 - 1)(\sqrt 3 +1)=2$.
Are there identities that would help with the construction for $ \sqrt[3]{2} $?
One I thought was useful in the first step to get [1; 3,...] was 
$ (\sqrt[3]{2}-1)( \sqrt[3]{4} + \sqrt[3]{2}+1 )=1$,
So you get:
$ \sqrt[3]{2}=1+( \sqrt[3]{2}-1 )=1+\frac {1}{ \sqrt[3]{4} + \sqrt[3]{2}+1 }= 1+\frac {1}{3+ (\sqrt[3]{4} + \sqrt[3]{2}-2)} $
Thanks for the help.
 [/INST] Starting from the column vector $(1,0,0,-2)$, consider the following steps:
Step a) Repeat multiplication by the matrix $A$
$$A=\begin{bmatrix}
1&0&0&0\\
3&1&0&0\\
3&2&1&0\\
1&1&1&1
\end{bmatrix}$$
while the coefficients of the resulting vector have different signs.
Step b) Reverse the coefficients of the vector, or equivalently multiply by
$$B=\begin{bmatrix}
0&0&0&1\\
0&0&1&0\\
0&1&0&0\\
1&0&0&0
\end{bmatrix}$$
Then the number of times you multiply by $A$ in step a gives the partial quotients of continued fraction of $\sqrt[3]{2}$.

For, starting from $(1,0,0,-2)$, successive multiplication by $A$ gives:
\begin{align}
(1,0,0,-2)
&\xrightarrow A\color{red}{(1,3,3,-1)}\\
&\xrightarrow A(1,6,12,6)
\end{align}
hence in step a we multiply by $A$ one time only, because $(1,6,12,6)$ have positive coefficients only, hence the first partial quotient is $1$:
$$\sqrt[3]{2}=1+\cdots$$
Apply step b to $(1,3,3,-1)$ we get $(-1,3,3,1)$.
Then applying step a to $(-1,3,3,1)$, successive multiplication by $A$ gives:
\begin{align}
(-1,3,3,1)
&\xrightarrow A(-1,0,6,6)\\
&\xrightarrow A(-1,-3,3,11)\\
&\xrightarrow A\color{red}{(-1,-6,-6,10)}\\
&\xrightarrow A(-1,-9,-21,-3)\\
\end{align}
hence the second partial quotient is $3$:
$$\sqrt[3]{2}=1+\frac 1{3+}\cdots$$
and so on...

This algorithm holds for every algebraic number of third degree which is the only positive root of it minimal polynomial.
For higher degree the matrix $A$ is enlarged as in the Tartaglia-Pascal triangle; for example for fourth degree:
$$A=\begin{bmatrix}
1&0&0&0&0\\
4&1&0&0&0\\
6&3&1&0&0\\
4&3&2&1&0\\
1&1&1&1&1
\end{bmatrix}$$

For the intuition behind this algorithm.
Then vector $(1,0,0,-2)$ corresponds to the polynomial $p(x)=x^3-2$.
Multyplication by $A$ corresponds to $p(x)\mapsto p(x+1)$, while revesing in step b corresponds to $p(x)\mapsto x^3p(1/x)$.
Finally Descartes's signs rule provide the stopping criterion in step a.
</s>