Prove using Mathematical Induction that $2^{3n}-3^n$ is divisible by $5$ for all $n≥1$. I did most of it but I stuck here I attached my working 
tell me if I did correct or not thanks
My working:

EDITED: I wrote the notes as TEX
Prove using induction that $2^{3n} - 3^n \mod{5} = 0$.
Statement is true for $n = 1$: 
$$2^{3 * 1} - 3^1 = 2^3 - 3 = 8 - 3 = 5$$
$$5 \mod{5} = 0$$
Now for $n = p$ and $n = p + 1$:
$$2^{3(k+1)} - 3{k + 1} = 2 * 2^p - 3$$
$$=2(5n + 3) - 3=10n + 6 - 3 = 10n+3$$

It's hard to read your handwriting but it looks like you have the right idea but were sloppy in your execution and made so distributive error.
Assume if $n=k$ then statement is true and 
$2^{3k} - 3^k = 5P$ for some integer $P$.  (Always a good idea to specify what a variable is whenever you introduce it.)
$2^{3k+1} -3^{k+1}= 2*2^{3k} - 3$.
Theres two errors here:  $2^{3(k+1)} \ne 2^{3k+1}$ and ... why did the $3^{k+1}$ turn into a $3$.
You should have:
$2^{3(k+1)} - 3^{k+1} = 2^{3k + 3} - 3^{k+1} = 8*2^{3k} - 3*3^{k}$
Can you go from there?
$8*2^{3k} - 3*3^{k} = 5*2^{3k} + 3*2^{3k} - 3*3^k = 5*2^{3k} + 3(2^{3k} - 3^k) = 5*2^{3k} + 3(5P) = 5(2^{3k} + 3P)$
