# How to Break Down a Cubic Difference or Sum

After you’ve checked to see if there’s a Greatest Common Factor (GCF) in a given polynomial and discovered it’s a binomial that isn’t a difference of squares, you should consider that it may be a difference or sum of cubes.

A *difference of cubes* sounds an awful lot like a difference of squares, but it factors quite differently. A difference of cubes is a binomial that is of the form (something)^{3} – (something else)^{3}. To factor any difference of cubes, you use the formula *a*^{3} – *b*^{3} = (*a* – *b*)(*a*^{2} + *ab* + *b*^{2}).

A *sum of cubes* is a binomial of the form: (something)^{3} + (something else)^{3}. When you recognize a sum of cubes *a*^{3} + *b*^{3}, it factors as (*a* + *b*)(*a*^{2} – *ab* + *b*^{2}).

For example, to factor 8*x*^{3} + 27, you first look for the GCF. You find none, so now you use the following steps:

Check to see if the expression is a difference of squares.

You want to consider the possibility because the expression has two terms, but the plus sign between the two terms quickly tells you that it isn’t a difference of squares.

Determine if you must use a sum or difference of cubes.

The plus sign tells you that it may be a sum of cubes, but that clue isn’t foolproof. Time for some trial and error: Try to rewrite the expression as the sum of cubes; if you try (2

*x*)^{3}+ (3)^{3}, you’ve found a winner.Break down the sum or difference of cubes by using the factoring shortcut.

Replace

*a*with 2*x*and*b*with 3. The formula becomes [(2*x*) + (3)] [(2*x*)^{2}– (2*x*)(3) + (3)^{2}].Simplify the factoring formula.

This example simplifies to (2

*x*+ 3)(4*x*^{2}– 6*x*+ 9).Check the factored polynomial to see if it will factor again.

You’re not done factoring until you’re done. Always look at the “leftovers” to see if they’ll factor again. Sometimes the binomial term may factor again as the difference of squares. However, the trinomial factor

*never*factors again.In this example, the binomial term 2

*x*+ 3 is a first-degree binomial (the exponent on the variable is 1) without a GCF, so it won’t factor again. Therefore, (2*x*+ 3)(4*x*^{2}– 6*x*+ 9) is your final answer.