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.
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.
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 $$(-a^3 + a^2)^6 =1$$
Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
divide both sides by $a^2$ to find
$$0=a^2-a-1+\dfrac1a+\dfrac1{a^2}=\left(a-\dfrac1a\right)^2-\left(a-\dfrac1a\right)+1=0$$
$\implies a-\dfrac1a=-w$ where $w$ is a complex cube root of unity
$\implies a^2=1-wa$
$\implies a^3=a(1-wa)=a-wa^2=a-w(1-wa)=a(1+w^2)-w=-wa-w$
Compare the values of $wa$
$$1-a^2=a^3-w$$
$$\iff a^2-a^3=1+w=-w^2\ne1$$
$$\implies(a^2-a^3)^2=?\ne1$$
$$\implies(a^2-a^3)^3=?\ne1$$
$$\implies(a^2-a^3)^6=?$$
