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

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Answer 1 · 2016-01-03T17:04:28

Using the identity $a^3-b^3=(a-b)*(a^2+ab+b^2)$ we have that

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

Answer 2 · 2016-01-03T17:07:47

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

This is a difference of cubes.

Differences of cubes can be factored as follows:

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

In your situation, $a=x$ and $b=1$ , since $a^3=x^3$ and $b^3=1$ .

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

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

How do you factor completely x^3 - 1x^3 - 1?