# Sum of Cubes

Reference > Mathematics > Algebra > Factoring Higher Degree PolynomialsIn the previous section, we learned how to factor a binomial in which the two terms are perfect cubes, and they are being subtracted. In this section, we'll learn how to factor a similar binomial, in which the two perfect cubes are being added. First, we'll introduce this topic by asking you to perform another polynomial multiplication:

(a + b)(a^{2} - ab + b^{2})

a(a^{2} - ab + b^{2}) +b(a^{2} - ab + b^{2})

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

a^{3} + b^{3}

And thus we have our factoring rule:

**Sum of Cubes**

a^{3} + b^{3} = (a + b)(a^{2} - ab + b^{2})

If you did the section on difference of cubes, you shouldn't need too many examples of how to use this rule, since it is almost identical to the Difference of Cubes rule. Here's one example:

**Example**

Factor 128x^{6} + 2y^{9}

Solution

First we note that 2 can be factored out of the entire expression:

128x^{6} + 2y^{9} = 2(64x^{6} + y^{9})

Since 64x^{6} and y^{9} are both perfect cubes, a = 4x^{2} and b = y^{3}

128x^{6} + 2y^{9} = 2(4x^{2} + y^{3})((4x^{2})^{2} - 4x^{2}(y^{3}) + (y^{3})^{2})

128x^{6} + 2y^{9} = 2(4x^{2} + y^{3})(16x^{4} - 4x^{2}y^{3} + y^{6})

As with the Difference of Cubes rule, we can do a cursory check of our answer by multiplying the first terms of the two polynomials, and multiplying the last terms of the two polynomials:

4x^{2}(16x^{4}) = 64x^{6}

y^{3}(y^{6}) = y^{9}

These match the two terms in the original binomial, which suggests that we're on the right track. It's still possible we have errors, and if you are uncertain, you can do a full check by completely multiplying out your polynomials to see if you get back to the original expression.

## Questions

^{3}+ 8y

^{3}

^{4}+ 10x

^{3r}+ y

^{6r}z

^{12r}

^{3}y

^{3}+ z

^{3}

^{3}+ 1

^{3})