<s>[INST] Using mathematical induction prove that $ 11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 Prove that $ \phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6  $ is divisible by 8 for all $n \in N$.
Base: $ n = 0 $
$ 8 | 11 + 3 - 6 $ is obvious.
Now let $\phi(n)$ be true we now prove that is also true for $ \phi(n+1)$.
So we get $ 11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8. 
For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here. 
 [/INST] Setup the same as your current work:
$\dots$
$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$
$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$
$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^n}_{\text{unknown}}$
Now, what can we say about $22\cdot 3^n+18\cdot 7^n$?  Anything?  You say in a previous example, you had to run a second induction proof to finish, might that be useful here?
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