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How to factorise $x^6 + 1

15 Apr 2013 — The fully factorized form of x6+1 (over the complex numbers) is therefore (x−i)(x+i)(x−12√3−12i)(x+12√3−12i)(x−12√3+12i)(x+12√3+12i).Read more

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Link: https://math.stackexchange.com/questions/362203/how-to-factorise-x6-1

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Factor x^6-1

Since both terms are perfect cubes, factor using the difference of cubes formula, a3−b3=(a−b)(a2+ab+b2) a 3 - b 3 = ( a - b ) ( a 2 + a b + b 2 ) where a=x2 a = ...Read more

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How to factorise $x^6 + 1

15 Apr 2013 — The fully factorized form of x6+1 (over the complex numbers) is therefore (x−i)(x+i)(x−12√3−12i)(x+12√3−12i)(x−12√3+12i)(x+12√3+12i).Read more

Tên miền: math.stackexchange.com Đọc thêm

[pre calc] how does x^6-1 produce or has anything to do ...

x6 - 1 = 0 has 6 solutions. Obviously 1 + 0i and -1 + 0i are two of them, but there are four more...Read more

Tên miền: reddit.com Đọc thêm

polynomials - Find all roots of $x^{6} + 1

$polynomials - Find all roots of $x^{6} + 1$

29 Mar 2014 — Hint: if x6=−1, then |x|6=1 and you can write x=cosθ+isinθ. details: Then the equation is, thanks to De Moivre theorem and cos2+sin2=1, ...Read more

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