2. Complex Numbers & Quadratics

Common Core Algebra 2 - 2020 Edition

2.02 Factoring with sum and difference of cubes

Lesson

Previously, factoring binomials only covered the difference of squares. Can a binomial be factored if the terms are cubic?

Polynomials may be factored in various ways, including, but not limited to grouping or applying general patterns such as difference of squares, sum and difference of cubes, and perfect square trinomials. Algebra I covered all of the main factoring methods except one: **sum and difference of cubes**.

A polynomial in the form $a^3+b^3$`a`3+`b`3 is a sum of cubes and in the form $a^3-b^3$`a`3−`b`3 is a difference of cubes.

Let's start by looking at the general forms of these rules.

General forms for factoring binomials with cubic terms

Sum of two cubes: $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$`a`3+`b`3=(`a`+`b`)(`a`2−`a``b`+`b`2)

Difference of two cubes: $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$`a`3−`b`3=(`a`−`b`)(`a`2+`a``b`+`b`2)

Careful!

The expression $a^3+b^3$`a`3+`b`3 is not the same as $\left(a+b\right)^3$(`a`+`b`)3.

$2^3+5^3$23+53 | $\ne$≠ | $\left(2+5\right)^3$(2+5)3 |

$8+125$8+125 | $\ne$≠ | $7^3$73 |

$133$133 | $\ne$≠ | $343$343 |

Now let's look at some examples and see this process in action.

Factor: $x^3+125$`x`3+125

Think: Since the expression is a sum of two terms and each are perfect cubes ($x$`x`and $5$5, respectively), then apply:

$a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$`a`3+`b`3=(`a`+`b`)(`a`2−`a``b`+`b`2)

Do: Using the perfect cubes, substitute the values into formula for $a$`a` and $b$`b`.

$a^3+b^3$a3+b3 |
$=$= | $\left(a+b\right)\left(a^2-ab+b^2\right)$(a+b)(a2−ab+b2) |

$x^3+5^3$x3+53 |
$=$= | $\left(x+5\right)\left(x^2-x\times5+5^2\right)$(x+5)(x2−x×5+52) |

$x^3+125$x3+125 |
$=$= | $\left(x+5\right)\left(x^2-5x+25\right)$(x+5)(x2−5x+25) |

Factor: $27x^3-1$27`x`3−1

Think: Since the expression is a difference of two terms and each are perfect cubes ($3x$3`x` and $1$1, respectively), then apply:

$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$`a`3−`b`3=(`a`−`b`)(`a`2+`a``b`+`b`2)

Do: Using the perfect cubes, substitute the values into formula for $a$`a` and $b$`b`.

$a^3-b^3$a3−b3 |
$=$= | $\left(a-b\right)\left(a^2+ab+b^2\right)$(a−b)(a2+ab+b2) |

$\left(3x\right)^3-1^3$(3x)3−13 |
$=$= | $\left(3x-1\right)\left(\left(3x\right)^2+3x\times1+1^2\right)$(3x−1)((3x)2+3x×1+12) |

$27x^3-1$27x3−1 |
$=$= | $\left(3x-1\right)\left(9x^2+3x+1\right)$(3x−1)(9x2+3x+1) |

Factor $x^3+512$`x`3+512.

Factor $4m^3-32n^3$4`m`3−32`n`3.

Interpret expressions that represent a quantity in terms of its context. ^Limited to polynomial and rational functions.

Interpret parts of an expression, such as terms, factors, and coefficients. ^Limited to polynomial and rational functions.

Interpret complicated expressions by viewing one or more of their parts as a single entity. ^Limited to polynomial and rational functions.

Use the structure of an expression to identify ways to rewrite it. ^Limited to polynomial and rational functions.