# 4. The Sum and Difference of Cubes

We came across these expressions earlier (in the section Special Products involving Cubes):

x3 + y3 = (x + y)(x2xy + y2) [Sum of two cubes]

x3y3 = (xy)(x2 + xy + y2) [Difference of 2 cubes]

Where do these come from? If you multiply out the right side of each, you'll get the left side of the equation.

Note: We cannot factor the right hand sides any further.

We use the above formulas to factor expressions involving cubes, as in the following example.

### Example

Factor 64x3 + 125

We use the Sum of 2 Cubes formula given above.

64x3 + 125

= (4x)3 + (5)3

= (4x + 5)[(4x)2 − (4x)(5) + (5)2]

= (4x + 5)(16x2 − 20x + 25)

As mentioned above, we cannot factor the expression in the second bracket any further. It looks like it could be factored to give (4x-5)2, however, when we expand this it gives:

(4x − 5)2 = 16x2 − 40x + 25

This "perfect square trinomial" is not the same as the expression we obtained when factoring the sum of 2 cubes.

### Exercises

Factor:

(1) x3 + 27

Using the Sum of 2 Cubes formula, we obtain:

x3 + 27

= (x)3 + (3)3

= (x + 3)[(x)2 − (x)(3) + (3)2]

= (x + 3)(x2 − 3x + 9)

(2) 3m3 − 81

Using the Difference of 2 Cubes formula, we obtain:

3m3 − 81

= 3(m3 − 27)

= 3(m3 − (3)3)

= 3(m − 3)[(m)2 + (m)(3) + (3)2]

= 3(m − 3)(m2 + 3m + 9)