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

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Answer 1 · 2016-01-03T17:04:28

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

$x^{3} - 1 = (x - 1) \cdot (x^{2} + x + 1)$ $x^3-1=(x-1)*(x^2+x+1)$

Answer 2 · 2016-01-03T17:07:47

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

This is a difference of cubes.

Differences of cubes can be factored as follows:

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

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

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

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

How do you factor completely x^3 - 1x3−1x^3 - 1?