# Sum or Difference of Odd Powers

Reference > Mathematics > Algebra > Factoring Higher Degree PolynomialsHaving learned about both sum of cubes and difference of cubes in this unit, we're now going to take those two rules and extend them to any odd power (for example, a^{5} - b^{5}, or a^{7} + b^{7}). These may look intimidating, but once you've learned sum of cubes and difference of cubes, these are actually fairly straightforward. Let's start by looking at the example above: a^{5} - b^{5}.

## Difference of Odd Powers

a^{5} - b^{5} = (a - b)(a^{4} + a^{3}b + a^{2}b^{2} + ab^{3} + b^{4})

This starts out just like a^{3} - b^{3}, with the binomial (a - b). What follows is an extended form of the difference of cubes trinomial; in this polynomial, we started with a^{4} and worked our way down by degrees, meanwhile working up to b^{4} by degrees. The thing that makes this simple to keep track of is that every term in the second factor has degree 4. So if we add our exponents for a and b in each term, we should get four.

**Example One**

Factor the following: 128x^{7} - 2187

**Solution**

In this problem, we note that both terms are perfect seventh powers, so we write out the following:

a^{7} - b^{7} = (a - b)(a^{6} + a^{5}b + a^{4}b^{2} + a^{3}b^{3} + a^{2}b^{4} + ab^{5} + b^{6})

Note that each term in the second factor has degree 6 (one less than 7).

We need to find a and b now. Since a^{7} = 128x^{7}, and b^{7} = 2187, a = 2x, and b = 3. The factorization is:

(2x - 3)((2x)^{6} + (2x)^{5}(3) + (2x)^{4}(3)^{2} + (2x)^{3}(3)^{3} + (2x)^{2}(3)^{4} + 2x(3)^{5} + 3^{6})

(2x - 3)(64x^{6} + 96x^{5} + 144x^{4} + 216x^{3} + 324x^{2} + 486x + 729)

## Sum of Odd Powers

You might be able to guess what the sum of odd powers factorization will look like. Here I'll show you the one for a^{7} + b^{7}.

a^{7} + b^{7} = (a + b)(a^{6} - a^{5}b + a^{4}b^{2} - a^{3}b^{3} + a^{2}b^{4} - ab^{5} + b^{6})

As you can see, the binomial has a plus sign (just like in the sum of cubes rule - which completely makes sense, since cubing is raising to an odd power!). The second factor is just like the second factor for Difference of Odd Powers, except that every other term is negative.

**Example Two**

Factor the following: 64x^{10} + 2y^{5}

**Solution**

First we note that 2 can be factored out of both terms, giving the following:

64x^{10} + 2y^{5} = 2(32x^{10} + y^{5})

Now we observe that both factors in the binomial are 5th powers. Thus, our factoring rule will be:

a^{5} + b^{5} = (a + b)(a^{4} - a^{3}b + a^{2}b^{2} - ab^{3} + b^{4})

Since a^{5} = 32x^{10} and b^{5} = y^{5}, a = 2x^{2}, and b = y. Now we plug these into our rule:

64x^{10} + 2y^{5} = 2(2x^{2} + b)((2x^{2})^{4} - (2x^{2})^{3}(y) + (2x^{2})^{2}(y)^{2} - (2x^{2})(y)^{3} + y^{4})

64x^{10} + 2y^{5} = 2(2x^{2} + b)(16x^{8} - 8x^{6}y + 4x^{4}y^{2} - 2x^{2}y^{3} + y^{4})

Note that, as with the previous example, our exponents make a nice pattern, as do our coefficients. If they didn't we ought to be concerned that we've done something wrong!

## Questions

^{5}- 1

^{8}- 384x

^{7}+ 2

^{5}+ y

^{10}z

^{10}