<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 that
$$
(a^2-a^3)^6-1=(a^6 - 2a^5 + a^4 - a^3 + a^2 + 1)(a^4 - a^3 - a^2 + a + 1)(a^3 - a^2 + 1)(a^3 - a^2 - 1)(a^2 - a + 1),
$$
so that the claim of the title follows.
</s>