Why is $(3, 1+\sqrt{-26})^3=( 1+\sqrt{-26})$ in $\mathbb Z[\sqrt{-26}]$? $a = 3, b = 1+\sqrt{-26}$ then $(a,b)^3=(a^3,b^3,a^2b,ab^2)$
each generator except $a^3$ has a $b$ factor and $\bar b b=27$, so $"\subseteq"$. Now the question is how to obtain $b$ using these generators. Is there a general formula, which states in which case this is possible or even is  there an algorithm. I made it very complicated I think;
$a^3 = 27,\quad$
$b^3 = -77+23\sqrt{-26},\quad$
$a^2b = 9+9\sqrt{-26},\quad$
$ab^2 = -75+6\sqrt{-26}$
$3*a^3+ab^2=81+(-75+6\sqrt{-26})=6+6\sqrt{-26}$
$a^2b-ans = 3+3\sqrt{-26}$
are these correct so far ?
 A: True by below, $ $ since here: $\,a\!=\!3,\,b+\bar b=\color{#0af}2\,\Rightarrow\, \color{#0a0}{(a,b,\bar b)} = (3,b,\bar b,\color{#0af}2)=1$.
Lemma $\, $ If $\  a^3 = b\bar b\ $ then $\ (a,b)^3 = (b)\iff  \color{#0a0}{(a,b,\bar b)}=1$
Proof $\ \ (a,b)^3\! = (b)((\bar b)\!+\!(a,b)^2) = (b)\!\iff\!$ $(\bar b)\!+\!(a,b)^2\! = 1\!\!\!\overset{\color{#c00}{\rm EL}\!\!}\iff\!$
$\color{#0a0}{(\bar b)\!+\!(a,b)}=1$
A: I think $b^3$ should be $-77 - 23\sqrt{-26}$ which makes $-(2a^3 + b^3) = 23(1+\sqrt{-26})$ and you have already found $6(1+\sqrt{-26})$ in the ideal. Together they will give you $1+\sqrt{-26}$.
I have not come across any general method for these.
A: To be explicit, you could use linear algebra:
$$\begin{align}
a^3&=27&&\mapsto\begin{bmatrix}27\\0\end{bmatrix}\\
b^3&=-77-23\sqrt{-26}&&\mapsto\begin{bmatrix}-77\\-23\end{bmatrix}\\
a^2b&=9+9\sqrt{-26}&&\mapsto\begin{bmatrix}9\\9\end{bmatrix}\\
ab^2&=-75+6\sqrt{-26}&&\mapsto\begin{bmatrix}-75\\6\end{bmatrix}\\
\end{align}$$
It would suffice to have an integer solution to the equation $$wa^3+xb^3+ya^2b+zab^2=b$$ The corresponding augmented matrix is
$$\begin{bmatrix}
27&-77&9&-75&1\\
0&-23&9&6&1
\end{bmatrix}$$
This reduces to
$$\begin{bmatrix}
23&0&-18&-81&-2\\
0&23&-9&-6&-1
\end{bmatrix}$$
You need choices for $y$ and $z$ such that $18y+81z-2$ and $9y+6z-1$ are each divisible by $23$. In other words, you want to solve
$$\begin{bmatrix}
18&81\\
9&6
\end{bmatrix}
\begin{bmatrix}
y\\
z
\end{bmatrix}
\equiv
\begin{bmatrix}
2\\
1
\end{bmatrix}$$
modulo $23$. So we can again use linear algebra (with some division mod $23$) to find
$$
\begin{bmatrix}
y\\
z
\end{bmatrix}
\equiv
\begin{bmatrix}
18&81\\
9&6
\end{bmatrix}^{-1}
\begin{bmatrix}
2\\
1
\end{bmatrix}
\equiv
\begin{bmatrix}
2\\
1
\end{bmatrix}$$
We can just go with $y=2$ and $z=1$. This gives $(w,x,y,z)=(5,1,2,1)$. So it seems that
$$5a^3+b^3+2a^2b+ab^2=b$$
