There are no 3×3 matrices over QQ of multiplicative order 8

Show that there do not exist $3 \times 3$ matrices $A$ over $\mathbb{Q}$ such that $A^8 = I$ and $A^4 \neq I$.

If $A$ is such a matrix, then the minimal polynomial of $A$ divides $x^8-1 = (x^4-1)(x^4+1)$ but not $x^4-1$; so the minimal polynomial of $A$ divides $x^4+1$. Since $A$ has dimension 3, the characteristic polynomial of $A$ has degree 3, so the minimal polynomial has degree at most 3. Note that $(x+1)^4+1 = x^4 + 4x^3 + 6x^2 + 4x + 2$ is Eisenstein at 2, and so $x^4+1$ is irreducible over $\mathbb{Q}$. (See Proposition 13 on page 309 in D&F, and Example 3 on page 310.) So we have a contradiction- no divisor of $x^4+1$ over $\mathbb{Q}$ can have degree between 1 and 3. So no such matrix $A$ exists.