# 4. The Sum and Difference of Cubes

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

x^{3}+y^{3}= (x+y)(x^{2}−xy+y^{2}) [Sum of two cubes]

x^{3}−y^{3}= (x−y)(x^{2}+xy+y^{2}) [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 64*x*^{3} + 125

**Answer:**

We use the Sum of 2 Cubes formula given above.

64*x*^{3} + 125

= (4

x)^{3}+ (5)^{3}= (4

x+ 5)[(4x)^{2}− (4x)(5) + (5)^{2}]= (4

x+ 5)(16x^{2}− 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 (4*x*-5)^{2}, however, when we expand this it gives:

(4

x− 5)^{2}= 16x^{2}− 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) ***x*^{3} + 27

Answer

Using the Sum of 2 Cubes formula, we obtain:

*x*^{3} + 27

= (

x)^{3}+ (3)^{3}= (

x+ 3)[(x)^{2}− (x)(3) + (3)^{2}]= (

x+ 3)(x^{2 }− 3x+ 9)

**(2) **3*m*^{3} − 81

Answer

Using the Difference of 2 Cubes formula, we obtain:

3*m*^{3} − 81

= 3(

m^{3}− 27)= 3(

m^{3}− (3)^{3})= 3(

m− 3)[(m)^{2}+ (m)(3) + (3)^{2}]= 3(

m− 3)(m^{2}+ 3m+ 9)