prove that $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 Prove that a number $2^n+2^{n-1}+2^{n-2}+8^n-8^{n-2}$ is a multiple of 7 for every natural $n\ge2$. I am not sure how to start.

This is relatively easy to solve even without noticing some clever factorization, simply by checking several possible cases.
It suffices to check what are remainder of powers of $2$ modulo $7$.
You can see that:
$2^0 \equiv 1 \pmod 7$
$2^1 \equiv 2 \pmod 7$
$2^2 \equiv 4 \pmod 7$
$2^3 \equiv 1 \pmod 7$
$2^4 \equiv 2 \pmod 7$
$2^5 \equiv 4 \pmod 7$
We see, that this repeats with period $3$. We get that $2^k\equiv 1\pmod 7$ if $k$ is multiple of $3$, $2^k\equiv 2\pmod7$ if the remainder modulo $3$ is $1$ and $2^k\equiv 4\pmod 7$ in the remaining case. 
Let us look at the case $n=3k+1$.
Then we get
$2^n = 2^{3k+1} \equiv 2 \pmod 7$
$2^{n-1} = 2^{3k} \equiv 1 \pmod 7$
$2^{n-2} = 2^{3(k-1)+1} \equiv 4 \pmod 7$
$8^n=2^{3n} \equiv 1 \pmod 7$
$8^{n-2} = 2^{3(n-2)} \equiv 1 \pmod 7$
(In fact, for the powers of $8$ it is easier to notice that $8^n \equiv 1^n \equiv 1 \pmod 7$.)
By summing the above congruences we get
$2^n+2^{n-1}+2^{n-2}+8^n - 8^{n-2} \equiv 2+1+4+1-1 \equiv 7 \equiv 0 \pmod 7$.
I leave remaining cases $n=3k$ and $n=3k+2$ for you. (And, as you have seen in another answer, if you think about this a bit, you should be able to come up with a solution where you don't need to try various cases.) 
