EDGE Math 8

Products Giving a Sum and Difference of Two Cubes

Consider the following equation in which a binomial is multiplied by a trinomial:

(a + b) (a^{2} - a b + b^{2})

= a^{3} - a^{2} b + a b^{2} + a^{2} b - a b^{2} + b^{3}

If you will combine like terms on the right-hand side of the equation, you will get the following:

(a + b) (a^{2} - a b + b^{2}) = a^{3} + b^{3}

Notice that the resulting product is a sum of two cubes. As an exercise, you can also try deriving the following equation, in which the product is a difference of two cubes:

(a - b) (a^{2} + a b + b^{2}) = a^{3} - b^{3}

Example

Multiply each of the following:

1. $(x + 4) (x^{2} - 4 x + 16)$

2. $(2 p - 3) (4 p^{2} + 6 p + 9)$

Solution

1. Notice that the factors are of the form $(a + b) (a^{2} - a b + b^{2}), where$ $a = x$ and $b = 4.$ Thus, the product should be a sum of two cubes.

(x + 4) (x^{2} - 4 x + 16)

= x^{3} + 4^{3} = x^{3} + 64

2. The factors are of the form $(a - b) (a^{2} + a b + b^{2}), where$ $a = 2 p$ and $b = 3.$ Thus, the product should be a difference of two cubes.

(2 p - 3) (4 p^{2} + 6 p + 9)

= {(2 p)}^{3} - 3^{3}

= 8 p^{3} - 27