Skip to main content
Search IntMath
Close

450+ Math Lessons written by Math Professors and Teachers

5 Million+ Students Helped Each Year

1200+ Articles Written by Math Educators and Enthusiasts

Simplifying and Teaching Math for Over 23 Years

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

Answer:

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

Answer

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

Answer

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)