2. Just as there is a special rule for factoring the difference of two squares, there are special rules for factoring the sum or difference of two cubes.

3. Example 3A: Factoring the Sum or Difference of Two Cubes
Factor the expression.
4x x
4x(x3 + 27)
Factor out the GCF, 4x.
4x(x3 + 33)
Rewrite as the sum of cubes.
Use the rule a3 + b3 = (a + b)  (a2 – ab + b2).
4x(x + 3)(x2 – x  )
4x(x + 3)(x2 – 3x + 9)

4. Example 3B: Factoring the Sum or Difference of Two Cubes
Factor the expression.
125d3 – 8
Rewrite as the difference of cubes.
(5d)3 – 23
(5d – 2)[(5d)2 + 5d  ]
Use the rule a3 – b3 = (a – b)  (a2 + ab + b2).
(5d – 2)(25d2 + 10d + 4)

5. Check It Out! Example 3a
Factor the expression.
8 + z6
Rewrite as the difference of cubes.
(2)3 + (z2)3
(2 + z2)[(2)2 – 2  z^2 + (z2)2]
Use the rule a3 + b3 = (a + b)  (a2 – ab + b2).
(2 + z2)(4 – 2z^2 + z4)

6. Check It Out! Example 3b
Factor the expression.
2x5 – 16x2
2x2(x3 – 8)
Factor out the GCF, 2x2.
Rewrite as the difference of cubes.
2x2(x3 – 23)
Use the rule a3 – b3 = (a – b)  (a2 + ab + b2).
2x2(x – 2)(x2 + x  )
2x2(x – 2)(x2 + 2x + 4)

7. Example 4: Geometry Application
The volume of a plastic storage box is modeled by the function V(x) = x3 + 6x2 + 3x – 10. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).
V(x) has three real zeros at x = –5, x = –2, and x = 1. If the model is accurate, the box will have no volume if x = –5, x = –2, or x = 1.

8. Example 4 Continued
One corresponding factor is (x – 1). 1
–10
Use synthetic division to factor the polynomial. 1 7 10 1 7 10
V(x)= (x – 1)(x2 + 7x + 10)
Write V(x) as a product.
V(x)= (x – 1)(x + 2)(x + 5)
Factor the quadratic.

9. Check It Out! Example 4
The volume of a rectangular prism is modeled by the function V(x) = x3 – 8x2 + 19x – 12, which is graphed below. Identify the values of x for which V(x) = 0, then use the graph to factor V(x).
V(x) has three real zeros at x = 1, x = 3, and x = 4. If the model is accurate, the box will have no volume if x = 1, x = 3, or x = 4.

10. Check It Out! Example 4 Continued
One corresponding factor is (x – 1). 1
1 –8 19 –12
Use synthetic division to factor the polynomial. 1 –7 12 1 –7 12
V(x)= (x – 1)(x2 – 7x + 12)
Write V(x) as a product.
V(x)= (x – 1)(x – 3)(x – 4)
Factor the quadratic.