Basic algebraic/arithmetic manipulations I do not know the rules governing the transformation of each member of the following group of equations into the the next one:
$$ 5(1 + 2^{k -1} + 3^{k -1}) - 6(1 + 2^{k -2} + 3^{k -2}) + 2 
\\= (5 -6 + 2) + (5 \times 2^{k - 1} - 6 \times 2^{k - 2}) + (5 \times 3^{k -1} - 6 \times 3^{k - 2})
\\= 1 + (5 - 6 \times 2^{-1}) \times 2^{k -1} + (5 - 6 \times 3^{-1}) \times 3^{k -1}
\\= 1 + (5 -3) \times 2^{k-1} + (5 - 2) \times 3^{k -1}
\\= 1 + 2 \times 2^{k -1} + 3 \times 3^{k -1}
\\= 1 + 2^{k} + 3^{k}$$
I need to learn rules such as these from scratch. Where can I learn them?
Could someone label the rule used to form each member of the group of equalities from the previous member?
 A: Here's an explanation of some of the lines:


*This line is obtained from line 1 by bringing the coefficients $5$, $6$ into their respective parentheses using the distributive law and then rearranging and regrouping terms using the commutative and associative laws. I assume there are no issues here, but if so, please ask.

*Factors $2^{k-1}$ and $3^{k-1}$ are pulled out of the 2nd and 3rd parentheses, again using the distributive law, but rules of exponents need to be used first. In the 2nd parentheses, the term $-6\cdot2^{k-2}$ has been rewritten as $-6\cdot2^{-1}\cdot2^{k-1}$ using the rule $a^{x+y}=a^xa^y$. Here $a=2$, $x=-1$ and $y=k-1$. In the 3rd parentheses, something similar was done: $-6\cdot3^{k-2}=-6\cdot3^{-1}\cdot3^{k-1}$.

*Arithmetic: $6\cdot 2^{-1}=6\cdot\frac{1}{2}=3$. This uses $a^{-x}=\frac{1}{a^x}$ with $a=2$ and $x=1$.

*Basic arithmetic.

*The law $a^xa^y=a^{x+y}$ is used again here: $2\cdot2^{k-1}=2^1\cdot2^{k-1}=2^{1+k-1}=2^k$ and similarly for the term with base $3$.

