4. The Sum and Difference of Cubes
We came across these expressions earlier (in the section Special Products involving Cubes):
x3 + y3 = (x + y)(x2 − xy + y2) [Sum of two cubes]
x3 − y3 = (x − y)(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)