Prove that $A,B$ are similar 
Let $A,B\in M_6(\mathbb{Q})$, such that the minimal polynomials are 
  $$m_A(x) = m_B(x) = x^2-x-1$$
Prove that above $\mathbb{Q}$, $A$ and $B$ are similar.

So $x^2-x-1$ is of course irreduable above $\mathbb{Q}$.
Hence, we can infer by the dimensions of $A,B$ (which are both $6$ as given) that the characteristic polynomials must be:
$$f_A(x) = f_B(x) = (x^2-x-1)^3$$
How should I continue? 
 A: I would read on https://en.wikipedia.org/wiki/Frobenius_normal_form 
The rough idea is (over any field), decompose $V$ into $T$-invariant subspaces $W$ which are cyclic, i.e. acting by powers of $T$ on a (generic) vector in $w \in W$ will generate a basis for $W$.  In particular, $w, Tw, T^2w, \ldots$ is eventually linearly dependent, and one obtains some "minimal" polynomial $p(T)w = 0$ for the subspace $W$, and this polynomial divides the minimal polynomial.  Then the change to this basis $w, Tw, \ldots$ is the rational canonical form.  Continue inductively.  Note that in the decomposition of $V$, at least one of the summands will have minimal polynomial equal to the minimal polynomial on all of $V$ (the minimal polynomial of $T$ on $V$ is the least common multiple of the minimal polynomials on the summands).
That your minimal polynomial is irreducible over $\mathbb{Q}$ means that the minimal polynomial restricted to each invariant cyclic subspace is also $x^2 - x - 1$ (since it must be a factor), which determines the Frobenius normal form.  So every matrix with the properties you describe have the same Frobenius normal form, so they're similar.
