﻿ Factoring Sum of Cubes the formula. Explained with 3 different rationales Unfortunately, in the last year, adblock has now begun disabling almost all images from loading on our site, which has lead to mathwarehouse becoming unusable for adlbock users.

# Factoring Sums of Cubes

### Quick Overview

• The Formula: $$a^3 + b^3$$ = $$(a + b)(a^2 - ab + b^2)$$
• This factoring works for any binomial that can be written as $$a^3 + b^3$$
• The discussion below includes 2 ways of demonstrating that this formula is valid. ### Explanation of the Formula -- Direct Method

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*}

### Explanation of the Formula -- Division Method

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 1

Find a solution to $$a^3 + b^3 = 0$$.

\begin{align*} a^3 + b^3 & = 0\\ a^3 & = -b^3\\ \sqrt{a^3} & = \sqrt{-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 2

Find 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{22mm}\\ & ab^2 + b^3 \hspace{15mm}\\ & - \underline{(ab^2 + b^3)} \hspace{13mm}\\ & 0 \hspace{13mm} \end{array} \end{align*}

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

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

### Example

Show that $$x^3 + 1$$ factors into $$(x + 1)(x^2 - x + 1)$$.

Step 1

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

\begin{align*} \blue{(x+1)}(x^2 - x + 1) & = x^2\blue{(x+1)} - x\blue{(x+1)} + 1\blue{(x+1)}\\ & = x^3 + x^2 - x^2 - x + x+1\\ & = x^3 +1 \end{align*}

Step 1 (Alternate Solution)

Show that $$(x + 1)(x^2 - x + 1)$$ matches the correct pattern for the formula.

Since we want to factor $$x^3 + 1$$, we first identify $$a$$ and $$b$$.

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

Likewise, since $$b$$ is the cube root of the second term, we know $$b = \sqrt 1 = 1$$.

Step 2

Write 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 + 1 & = (\blue x + \red 1)(\blue x^2 - \blue x\cdot \red 1 + \red 1^2)\\ & = (x+1)(x^2 - x + 1) \end{align*}

$$x^3 + 1 = (x+1)(x^2 - x +1)$$ 