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.
Let's start by looking at the general forms of these rules.
Sum of two cubes: $a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$a3+b3=(a+b)(a2−ab+b2)
Difference of two cubes: $a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$a3−b3=(a−b)(a2+ab+b2)
The expression $a^3+b^3$a3+b3 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: $x^3+125$x3+125
Think: Since the expression is a sum of two terms and each are perfect cubes ($x$xand $5$5, respectively), then apply:
$a^3+b^3=\left(a+b\right)\left(a^2-ab+b^2\right)$a3+b3=(a+b)(a2−ab+b2)
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$27x3−1
Think: Since the expression is a difference of two terms and each are perfect cubes ($3x$3x and $1$1, respectively), then apply:
$a^3-b^3=\left(a-b\right)\left(a^2+ab+b^2\right)$a3−b3=(a−b)(a2+ab+b2)
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 $x^3+512$x3+512.
Factor $4m^3-32n^3$4m3−32n3.