### Quick Overview

- $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
- This factoring works for any binomial that can be written as $$a^3 - b^3$$.
- The discussion below includes three separate ways of demonstrating that this formula is valid.

- $$a^3 - b^3 = (a - b)(a^2 + ab + b^2)$$
- This factoring works for any binomial that can be written as $$a^3 - b^3$$.
- The discussion below includes three separate ways of demonstrating that this formula is valid.

On the page about the sum of cubes we showed that $$a^3 + b^3 = a^2 - ab + b^2$$. We can use this formula to find a factorization for $$a^3 - b^3$$.

We start by writing $$a^3 - b^3$$ as $$a^3 + (-b)^3$$ and then using the sum of cubes pattern.

$$ \begin{align*} a^3 - b^3 & = a^3 + (-b)^3\\ & = \left(a + (-b)\right)\left(a^2 + a(-b) + (-b)^2\right)\\ & = (a - b)(a^2 - ab + b^2) \end{align*} $$

We could also determine the validity of the formula using the same methods we established in the sum of cubes lesson.

We can verify the factoring formula by expanding the result and seeing that it simplifies to the original, as follows.

$$ \begin{align*} \blue{(a-b)}(a^2 + ab + b^2) & = a^2\blue{(a-b)} + ab\blue{(a-b)} + b^2\blue{(a-b)}\\ & = a^3 - a^2b + a^2b - ab^2 + ab^2 - b^3\\ & = a^3 \blue{- a^2b + a^2b}\,\,\red{ - ab^2 + ab^2} - b^3\\ & = a^3 + \blue 0 + \red 0 - b^3\\ & = a^3 - b^3 \end{align*} $$

Another way to confirm to the formula is to find a solution to $$a^3 - b^3 = 0$$, and then use division to find the factored form.

Step 1Find a solution to $$a^3 - b^3 = 0$$.

$$ \begin{align*} a^3 - b^3 & = 0\\ a^3 & = b^3\\ \sqrt[3]{a^3} & = \sqrt[3]{b^3}\\ a & = b \end{align*} $$

One of the solutions is $$a = b$$. Adding $$b$$ to both sides of this equation gives us $$a - b = 0$$, which means $$(a - b)$$ is a factor of $$a^3 - b^3$$.

Step 2Find the other factor of $$a^3 - b^3$$ using polynomial division.

$$ \begin{align*} \begin{array}{rrcrcrcr} & a^2 \hspace{2mm} +ab \hspace{2mm} +b^2 \hspace{22mm} \\ & {a - b\,\,} \enclose{longdiv}{a^3 + 0\, a^2b + 0\, ab^2 - b^3} \hspace{10mm} \\ & \underline{-(a^3 - a^2b)} \hspace{35mm} \\ & a^2b + 0\,ab^2 - b^3\ \hspace{12mm} \\ & - \underline{(a^2b - ab^2)} \hspace{23mm}\\ & ab^2 - b^3 \hspace{15mm} \\ & - \underline{ab^2 - b^3)} \hspace{13mm}\\ & 0 \hspace{15mm} \end{array} \end{align*} $$

Since $$a - b$$ divides evenly into $$a^3 - b^3$$, we know

$$ (a - b)(a^2 + ab + b^2) = a^3 - b^3 $$

Show that $$x^3 - 27$$ factors into $$(x - 3)(x^2 + 3x + 9)$$.

Step 1Show that expanding $$(x - 3)(x^2 + 3x + 9)$$ results in $$x^3 - 27$$.

$$ \begin{align*} \blue{(x - 3)}(x^2 + 3x + 9) & = x^2\blue{(x - 3)} + 3x\blue{(x - 3)} + 9\blue{(x - 3)}\\ & = x^3 - 3x^2 + 3x^2 - 9x + 9x - 27\\ & = x^3 - 27 \end{align*} $$

Step 1 (Alternate Solution)Show that $$(x - 3)(x^2 + 3x + 9)$$ matches the correct pattern for the formula.

Since we want to factor $$x^3 - 27$$, we first identify $$a$$ and $$b$$.

Since $$a$$ is the cube root of the first term, we know $$a = \sqrt[3]{x^3} = x$$.

Likewise, since $$b$$ is the cube root of the second term, we know $$b = \sqrt[3]{27} = 3$$.

Step 2Write down the factored form.

$$ \begin{align*} a^3 + b^3 & = (\blue a - \red b)(\blue a^2 + \blue a \red b + \red b^2)\\ x^3 - 27 & = (\blue x - \red 3)(\blue x^2 + \blue x\cdot \red 3 + \red 3^2)\\ & = (x-3)(x^2 + 3x + 9) \end{align*} $$

Answer$$ x^3 - 27 = (x-3)(x^2 + 3x + 9) $$