Factorize algebraic equation and describe the steps taken.

We could write

$(a+3)^2(b+1)-2(a+3)(b+1)$

as

$(a+3)(a+3)(b+1)-2(a+3)(b+1)$

We do this because we notice a factor of $(a+3)(b+1)$ is repeated, so we can factorize it:

$(a+3)\textbf{(a+3)(b+1)}-2\textbf{(a+3)(b+1)} =\textbf{(a+3)(b+1)} [(a+3) - 2]$

Next, we can simplify the expression in the square brackets:

$\textbf{(a+3)(b+1)} [(a+3) - 2] = \textbf{(a+3)(b+1)} [a+1]$

And this is the result you obtained before.

Edit:

We want to understand the expression $(a+3) - 2$.

We must notice there is no multiplication involved here. For example, we can consider

$(-1)+3 = 2$
$(-1+3) = 2$
$-1+(3) = 2$
$-1 + 3 = 2$

Although the parenthesis can be uncomfortable to see, they don't mean that a multiplication takes place. They are simply used to separate numbers, but they are rather useless.

In our case,

$(a+3) - 2 = a + 3 - 2 = a +1$

The parentheses around $(a+3)$ mean nothing, and we can completely remove them.

IF we had something like $(a+3)2$, THEN we would take that to mean that $a+3$ is being multiplied by $2$. But in our case, we have $(a+3)-2$, which means that we must subtract $2$ from the number $a+3$.

Edit 2:
Parentheses matter when a minus sign is in front, when a number is in front or behind, when the multiplication sign ($*$ or $\times$) is used, or when the division sign is used. Examples:

$-(a+1) = -a - 1$
$2(a+1) = 2a +2$
$(a+1)3 = 3a + 3$
$2*(a+1) = 2a + 2$
$2\times(a+1) = 2a +2$

$2/(a+1) =\dfrac{2}{a+1}$

$(a+1)/2 = \dfrac{a+1}{2}$

Parentheses don't matter when they are being summed or subtracted something:

$(a+1)+2= a+3$
$(a+1)-2= a-1$

One has to be careful,though, because $2 - (a+1) =2-a-1 = 1-a$.