## Sum or Difference of Cubes

A sum of cubes or difference of cubes have two terms each of which are perfect cubes. That is, they can be cube rooted “nicely”.

$1, 8, 27, 64, 125, ldots$ are perfect squares, as well as polynomials like $x^3, y^3, x^6, y^9, a^3b^6,$ etc.

Unlike a difference of squares, there can be a plus or minus sign in between the two terms.

##### How to Factor a Sum or Difference of Squares

This is more easily explained using examples but there are some “lyrics to help you remember how to factor a sum or difference of cubes:

### Cube root the first term, keep the sign, cube root the second term

in order to make the first factor. Then, use the first factor to:

### Square the first term, negative product, square the last term.

Here are two basic examples: $$x^3+y^3=(x+y)(x^2-xy+y^2)$$ $$x^3-y^3=(x-y)(x^2+xy+y^2)$$

For example: $$x^3-8$$

The cube root of the first term is $x$. The cube root of the second term is 2. Therefore:$$x^3-8=(x-2)(x^2+2x+4)$$

### Practice

EXAMPLE 1

Factor: $45x^2-500$

###### SOLUTION

Using the Law of Division: $$2^{11} div 2^8=2^{11-8}=2^3$$

Evaluating: $2^3=8$

EXAMPLE 2

Factor: $25a^6-36b^4$

###### SOLUTION

Using the Law of Powers: $$(m^0)^{10}=m^0$$

We should more properly use the Law of Zero Exponents to write: $m^0=1$

EXAMPLE 3

Factor: $x^2+49$

###### SOLUTION

Since we are multiplying the two powers, we use the Law of Multiplication: $$x+4=7$$

Therefore: $x=3$

EXAMPLE 4

Factor: $x^4-16$

###### SOLUTION

Since we are multiplying the two powers, we use the Law of Multiplication: $$x+4=7$$

Therefore: $x=3$