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. 

Suppose $11*3^n + 3*7^n - 6 = 8k$
The $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$
$=3(11*3^n + 3*7^n-2) + 4*3*7^n $
$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$
$= 3(8k) + 4(3*7^n + 3)$;  $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.
$= 3(8k) + 8(\frac{3*7^n + 3}2) = 8(3k +  \frac{3*7^n + 3}2)$.
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Actually I like and am inspired by Bill Dubuques answer.
We want to prove $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 0 \mod 8$
And we know $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \mod 8$.
So it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \equiv 6 \mod 8$.
And if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \equiv 3^n + 3(-1)^{n}= f(n) \mod 8$.
So it's now just a matter of showing for $f(0) \equiv f(1) \equiv 6 \mod 8$.
Which is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$
