How do you factor $8x^3 - 1$ ?

Question

46121 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-04-09T03:55:49

Remember this formula for factorizing difference of 2 cubes:
$a^3-b^3 = (a-b)(a^2+ab+b^2)$

In $8x^3-1$ ,

$a^3=8x^3$
$b^3=1$

$a= root3(8x^3)$ = $2x$
$b=root3(1)$ = $1$

Substitute $a=2x$ , $b=1$ into the formula of $(a-b)(a^2+ab+b^2)$

$(2x-1)$ $((2x)^2$ + (1x $2x$ ) + $1^2$ $)$ = $(2x-1)$ $(4x^2$ + $2x$ + $1)$

$(2x-1)$ $(4x^2$ + $2x$ + $1)$ is the factorized form of $8x^3-1$

How do you factor 8x^3 - 18x^3 - 1?