How do you factor ab(a+b)+bc(b+c)+ca(c+a)+3abc?

In order to factor this expression, first open all parenthesis by distributing the term in front of the parenthesis. The given expression will then become

`a^2b + ab^2 + b^2c +bc^2 + ac^2 + a^2c + 3abc` .

We need to group these terms in a way so that we could find some common factors. Now there are 7 terms in the expression, which is not a very convenient number. However, the coefficient 3 in front of the abc term gives us a clue that this term should be broken up into 3: abc + abc + abc.

Now we have 9 terms, which we can group into three groups of three so that each group has a common factor:

`a^2b + a^2c + abc = a(ab + ac + bc)`

`ab^2 + b^2c + abc = b(ab + bc + ac)`

`bc^2 + ac^2 + abc = c(bc + ac + ab)`

Note that the expressions in parenthesis are the same, so the three groups of terms have a common trinomial factor! It can be factored out, and the entire sum of nine (originally, seven) terms becomes

`(a + b+c)(ab + ac + bc)` .

The given expression can be factored as `(a + b + c)(ab + ac + bc)`