<s>[INST] Let $a^4 - a^3 - a^2 + a + 1 = 0$ show that $(-a^3 + a^2)^6 = 1$ Hopefully I am reading the correct line from LMFDB.
Let $a$ be the algebraic number solving $a^4 - a^3  - a^2 + a + 1 = 0$, and consider the field extension generated by this polynomial $F =\mathbb{Q}(a) \simeq \mathbb{Q}[x]/(x^4 - x^3  - x^2 + x + 1)$.
Show that $(-a^3 + a^2)^6 = 1$ and $(-a^3 + a^2)^m \neq 0$ for $m < 6$ that is $-a^3 + a^2 \in \mathcal{O}_F$ is an element of the ring of order integers and is a unit of order six.
 [/INST] Note
\begin{align}
\frac{(-a^3 + a^2)^6 - 1 }{(-a^3 + a^2)^3 - 1  }&
=(-a^3 + a^2)^3 +1 = (-a^3 + a^2 +1 )[  (a^3 -a^2)^2 + (a^3 - a^2)+ 1]\tag1\\
\end{align}
where
\begin{align}
&  (a^3 -a^2)^2 +(a^3 - a^2)+ 1 \\
= & (a^3 -a^2+a-a)^2 +(a^3 - a^2+a -a)+ 1 \\
=& (a^3 -a^2+a)^2 -2a (a^3 -a^2+a) +(a^3 - a^2+a ) +a^2-a+ 1 \\
=& (a^2 -a+1)(a^4-a^3-a^2+a+1)=0
\end{align}
Plug into (1) to obtain
$$(-a^3 + a^2)^6 =1$$
</s>