Given matrices $A$ and $B$, prove that $A^{n}-B^{n}=\frac{1}{2}(7^{n}-5^{n})(A-B)$ $$\text{Given}$$
$$A=\begin{pmatrix} 2 & 5 \\  -3 & 10 \end{pmatrix}$$
$$\text{and}$$
$$B=\begin{pmatrix} 3 & -2 \\  4 & 9 \end{pmatrix}$$
$$\text{prove that}$$
$$A^{n}-B^{n}=\frac{1}{2}(7^{n}-5^{n})(A-B) \  \forall n \in \mathbb{N}$$
The first thing that I tried was to look for a pattern for $A^n$ and $B^{n}$, but, as I expected, I haven't been able to find any. Then I tried induction, and I think that I found a correct solution, but it is a very tedious one. Here's what I did:
$$\text{Base case:}$$
$$p(1):A-B=\frac{1}{2}(7-5)(A-B) - true$$
$$\text{suppose true}$$
$$p(k): A^{k}-B^{k}=\frac{1}{2}(7^{k}-5^{k})(A-B)$$
$$\text{then we need to prove}$$
$$p(k+1): A^{k+1}-B^{k+1}=\frac{1}{2}(7^{k+1}-5^{k+1})(A-B)$$
Then I multiplied the equation from $p(k)$ by $B^k$, and I subtracted it from the $p(k+1)$ equation to get rid of the $B^k$. My idea was to try and isolate $A^k$, and then prove by induction that the formula is, indeed correct, which would finish the proof. But as I said, this is a very tedious solution, and if someone could help me find a more intuitive and straight-forward one, I would appreciate that very much. Thanks!

Both $A$ and $B$ have characteristic polynomial $P=X^2-12X+35=(X-5)(X-7)$, which then annihilates both $A$ and B$.
Long division of polynomials yields $X^n=Q(X)P(X)+a_nX+b_n$, for some $a_n,b_n\in\mathbb{R}$.
Since $P(5)=P(7)=0$, we get $5^n=5a_n+b_n$ and $7^n=7a_n+b_n$.
Hence, $a_n=\dfrac{7^n-5^n}{2}$.
Note that we do not need to compute $b_n$. Since $P(A)=P(B)=0$, we get $A^n=a_nA+b_n$ and $B^n=a_n B+b_n$, so $A^n-B^n=a_n(A-B)$, which is exactly what you want.
