Question 44910
I tried to factor out an a from the {{{3a^3+3ab^2}}} part of the expression
and a b from the remaining part of the expression.  I was able to factor and a from the first expression because an a is common in each term of the expression.  That is, there is an a and {{{3a^3}}} and an a in {{{3ab^2}}}.  {{{3a^3}}} and {{{3ab^2}}} are called "terms" of the "expression" {{{3a^3+3ab^2}}}  Because they both have an a in them, I can take an a out.  Similar reasoning with b takes place too.  Using this reasoning, I obtained the expression.

{{{a(3a^2+3b^2)+b(2a^2+2b^2)}}}

Now I just think of factor by grouping to be a special name for this type of factoring situation.  However, it's the same exact idea as before.  Treat {{{a^2(3a+2b)}}} as one term and {{{b(2a^2+2b^2)}}} as another term of the expression {{{a(3a^2+3b^2) +b(2a^2+2b^2)}}}.  Unfortunately, there isn't a common factor in either of these terms.  So we will have to try another approach.

Let's start over.  Let's try to rearrange the terms so that we can get some terms with a common factor between them.  

{{{3a^3+2a^2b+3ab^2+2b^3}}}  

Using similar reasoning to how I factored before I obtain:

{{{a^2(3a+2b)+b^2(3a+2b)}}}

There is a common factor between each term, that is (3a+2b) is the common factor.  Thus, using factor by grouping I obtain the factored form:

{{{(3a+2b)(a^2+b^2)}}}