How do you factor completely $x^{6} - 1$ $x^6-1$ ?

Question

How do you factor completely $x^{6} - 1$ $x^6-1$ ?

1 Answer

Impact of this question

1282 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-05-13T11:58:19

$(x - 1) (x + 1) (x^{2} - x + 1) (x^{2} + x + 1)$ $(x-1)(x+1)(x^2-x+1)(x^2+x+1)$
$= (x - 1) (x + 1) (x - \frac{1}{2} - i \frac{\sqrt{3}}{2}) (x - \frac{1}{2} + i \frac{\sqrt{3}}{2}) (x + \frac{1}{2} - i \frac{\sqrt{3}}{2}) (x + \frac{1}{2} + i \frac{\sqrt{3}}{2})$ $=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)(x+1/2-i sqrt3/2)(x+1/2+ i sqrt3/2)$

Explanation:

The answer in the given form can be obtained directly by observing that it is a cubic in $x^{2}$ $x^2$ .

$x^{6} - 1 = {(x^{2})}^{3} - 1^{3}$ $x^6-1=(x^2)^3-1^3$

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

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

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

$= (x - 1) (x + 1) (x - \frac{1}{2} - i \frac{\sqrt{3}}{2}) (x - \frac{1}{2} + i \frac{\sqrt{3}}{2})$ $=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)$

$(x + \frac{1}{2} - i \frac{\sqrt{3}}{2}) (x + \frac{1}{2} + i \frac{\sqrt{3}}{2})$ $(x+1/2-i sqrt3/2)(x+1/2+ isqrt3/2)$

The expanded form can be directly obtained from

$(x^{6} - 1) = \prod (x - e^{i \frac{2 k π}{6}})$ $(x^6-1)=prod(x-e^(i(2kpi)/6))$ , k=0, 1, 2,..,5.,

using the six 6th roots of 1, ${e^{i \frac{2 k π}{6}}}$ ${e^(i(2kpi)/6)}$ , k=0, 1, 2,..,5.,

Science

Math

Humanities

... and beyond

How do you factor completely $x^{6} - 1$ $x^6-1$ ?

1 Answer

Explanation:

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

How do you factor completely x6−1x^6-1?

1 Answer

Explanation:

Related questions

Impact of this question

How do you factor completely $x^{6} - 1$ $x^6-1$ ?