Applying rules of algebra when working with multiplication and exponents I'm taking an online course and help is hard to find.  This specific problem has to do with recurrence relation.  I apologize for being too general but I'm just looking for help in how to go about solving this problem.
The entire problem is:

Let $d_0,d_1,d_2,\ldots$ be defined by the formula $d_n=3^n-2^n$ for all integers $n\ge0$.  Show that this sequence satisfies the recurrence relation:  $d_k=5d_{k-1}-6d_{k-2}$.

The step I can do without any trouble is finding the statements that represent the values we're dealing with in the relation, $d_{k-1}$ and $d_{k-2}$:
$d_k=3^k-2^k$
$d_{k-1}=3^{k-1}-2^{k-1}$
$d_{k-2}=3^{k-2}-2^{k-2}$
However, when it comes time to plugging that into the relation and simplifying it down to the original definition of $d_k=3^k-2^k$ I fail miserably.
$d_k=5d_{k-1}-6d_{k-2}$
$=5\left(3^{k-1}-2^{k-1}\right)-6\left(3^{k-2}-2^{k-2}\right)$
$=5\cdot3^{k-1}-5\cdot2^{k-1}-6\cdot3^{k-2}+6\cdot2^{k-2}$
$=5\cdot\frac{3^k}{3}-6\cdot\frac{3^k}{3^2}-5\cdot\frac{2^k}{2}+6\cdot\frac{2^k}{2^2}$
Is that a decent start?  Is there a better way to go, like breaking everything down into their simplest components, like so:
$d_k=5d_{k-1}-6d_{k-2}$
$=(2+3)\left(3^{k-1}-2^{k-1}\right)-(2\cdot3)\left(3^{k-2}-2^{k-2}\right)$
$=2\cdot\frac{3^k}{3}-2\cdot\frac{2^k}{2}+3\cdot\frac{3^k}{3}-3\cdot\frac{2^k}{2}-\left(\left(2\cdot\frac{3^k}{3^2}-2\cdot\frac{2^k}{2^2}\right)-\left(3\cdot\frac{3^k}{3^2}-3\cdot\frac{2^k}{2^2}\right)\right)$
But then what?
If I go either of these routes, I get stuck.  I feel I have two issues: a.) identifying the way to go that seem the most logical; and, b.) working towards a solution.  I don't know if I missed a big chunk in Algebra or if my brain just doesn't see what's going on.  
What am I missing?  Are either of these steps valid things to try?  What are some general rules to follow to work these out?
Also, what specific discipline of Algebra is this?  I don't think my course is introducing us to this stuff.  I think it assumes we already know how to work these out.

I start from what you wrote down:
$$5d_{k-1}-6d_{k-2}=5\cdot3^{k-1}-6\cdot3^{k-2}-5\cdot2^{k-1}+6\cdot2^{k-2}$$
and I want to get to $d_k=3^k-2^k$.
I rewrite $3^{k-1}=3\cdot3^{k-2}$, and similarly for $2^{k-1}$:
$$5\cdot3\cdot3^{k-2}-6\cdot3^{k-2}-5\cdot2\cdot2^{k-2}+6\cdot2^{k-2}$$
$$=15\cdot3^{k-2}-6\cdot3^{k-2}-10\cdot2^{k-2}+6\cdot2^{k-2}$$
I collect $3^{k-2}$ and $2^{k-2}$:
$$=(15-6)3^{k-2}-(10-6)2^{k-2}=9\cdot3^{k-2}-4\cdot2^{k-2}=3^2\cdot3^{k-2}-2^2\cdot2^{k-2}$$
From here I collapse the exponents to get the desired result.
$$=3^k-2^k$$
