Factoring Sum and Difference of Cubes

**magentarita** · August 6th 2008, 12:34 PM

There are so many different lectures on how to factor the sum and difference of cubes.

What is the easiest method to use?

For example:

How do you factor the following two questions:

(1) x^6 - 1

(2) x^3 - 125

**skeeter** · August 6th 2008, 12:48 PM

so many? only two I know of ...

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

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

$x^6 - 1 = (x^2)^3 - 1^3$ see the "a" and "b" ?

$x^3 - 125 = x^3 - 5^3$ ditto?

now use the factoring pattern for
$a^3 - b^3$ .

**Soroban** · August 6th 2008, 02:37 PM

Hello, magentarita!

There is a trick for memorizing the cube-factoring.

There are two forms:

. . $\begin{array}{cccc}\text{Sum of cubes:} & a^3+ b^3 &=& (a+b)\,(a^2-ab+ b^2) \\<br /> \text{Diff. of cubes:} & a^3-b^3 &=& (a-b)\,(a^2 + ab + b^2) \end{array}$

They can be combined: . $a^3 \pm b^3 \;=\;(a \pm b)\,(a^2 \mp ab + b^2)$

We must memorize the letters: . $a^3 \qquad b^3 \;\;=\;\;(a\qquad b)\,(a^2 \qquad ab \qquad b^2)$

To place the signs, remember the word SOAP:

. . . . . . . . . . . . . . .Opposite
. . . . . . . . . . . . . . . . . $\downarrow$
. . $a^3 \;{\color{red}\pm} \;b^3 \;\;=\;\; (a \;{\color{red}\pm} \;b)\,(a^2 \;{\color{red}\mp} \;ab \;{\color{red}+} \;b^2)$
. . . . . . . . . . . . $\uparrow\qquad \qquad \quad\;\:\uparrow$
. . . . . . . . . . Same . . . . Always Positive

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