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I was reading an example where the purpose was to compute a certain Galois group. Along the way, the writer says : note $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$. But how do I note this? I understand you can factorize by $x^2-1$, since when I draw on the unit circle I see that $-1$ and $+1$ are roots. But for the rest?

Edit :

I see you can then factor $(x^2-1)(x^4+x^2+1)$ and than substitute $x^2=y$ and solve quadratic equation but can you actually see the solution visually?

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  • $\begingroup$ You note it by .... doing it. Just multiply it out. It doesn't matter what steps you take to do it. If you do the arithmetic correctly you get that answer. $\endgroup$
    – fleablood
    Jan 17, 2019 at 23:22
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    $\begingroup$ @fleablood he doesn't want to check the factorization once he already knows it (that's the easy part) - he wants to know how to factorize the LHS without knowing the RHS $\endgroup$ Jan 17, 2019 at 23:24
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    $\begingroup$ Usually a writer says "Note WWWW" precisely because it is not clear that every reader will know WWWW. Then enough information is given for the reader to confirm WWWW, but not necessarily enough information for any particular reader to reconstruct WWWW from the ground floor. $\endgroup$
    – Will Jagy
    Jan 17, 2019 at 23:38
  • $\begingroup$ $$x^n-1 = \prod_{d\mid n}\Phi_d(x) $$ $\endgroup$ Jan 18, 2019 at 9:42

3 Answers 3

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$x^6-1=(x^3+1)(x^3-1)\\x^3-1=(x-1)(x^2+x+1)\\x^3+1=(x+1)(x^2-x+1)$
might help.

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    $\begingroup$ Nice! Interestingly, you can view the original expression as a difference of squares first, then the resulting factors as a sum or difference of cubes, or you can do it the other way around and use a difference of cubes first. Both work $\endgroup$ Jan 17, 2019 at 23:23
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Also, by using your idea we obtain: $$x^6-1=(x^2-1)(x^4+x^2+1)=(x^2-1)(x^4+2x^2+1-x^2)=$$ $$=(x^2-1)((x^2+1)^2-x^2)=(x^2-1)(x^2-x+1)(x^2+x+1).$$

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$x^6 - 1 = (x^2 -1)(x^4 +x^2 + 1)$ but

$x^6 - 1 = (x^3 -1)(x^3 + 1)$.

So $(x^4 +x^2+1)$ can be factored further. As can $x^3 -1$ and $x^3 +1$.

Factoring the obvious $(x^2 -1) = (x+1)(x-1)$ and $(x^3 -1) = (x-1)(x^2 + x + 1)$ we get:

$x^6 - 1 = (x+1)(x-1)(x^4 +x^2 + 1)$ but

$x^6 - 1 = (x-1)(x^2 + x+1)(x^3 + 1)$

So we know that $x+1$ factors either $x^3 + 1$ or $x^2 + x+1$ and the resulting factors will fact $x^4 + x^2 + 1$.

Hopefully it is clear $x^3 + 1 = (x+1)(x^2 -x +1)$

And that gives us our result:

$x^6 -1 = (x^3 -1)(x^3 +1)=$

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

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

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

.....

Slick thing that only becomes apparent in hindsight:

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

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

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

====Old answer====

You note it by doing it:

$(x^2−1)(x^2+x+1)(x^2−x+1) =... = x^6 -1$.

It doesn't matter which steps you take you will get that result.

But if I were looking for short slick answers I would do:

$x^6 -1 =(x^2-1)(x^4 + x^2 + 1)=(x-1)(x+1)(x^4+x^2 + 1)$

Now admittedly $x^4 + x^2 + 1$ isn't immediately obvious how to factor but knowing

$x^6 -1 = (x^3 -1)(x^3 + 1)=(x-1)(x^2 + x + 1)(x^3+1)$

AND

$x^6 -1 = (x^2 -1)(x^4 + x^2 + 1) = (x-1)(x+1)(x^4+x^2+1)$

we know they must factor down.

So dividing $x^3 + 1$ by $x+1$ we get.

$x^6 -1 = (x^3 -1)(x^3 + 1)=$

$(x-1)(x^2 + x + 1)(x+1)(x^2 -x + 1)$ and.... that's it.

$x^6 -1 = (x-1)(x+1)(x^2 +x +1)(x^2 - x + 1)=$

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

This further means $(x^2 +x + 1)(x^2 -x +1) =x^4 + x^2 + 1$.

.....

OR

....

Going the other way:

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

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

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

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

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

$x^6 + x^4 + x^2 - x^4 -x^2 -1 =$

$x^6 -1$.

....

But, seriously, when they say "note that..." you note by ... observing it's simply a fact. It's true.

Even if I didn't see any clever way to do it, I could still.... just do it.

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

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

$(x^4 + x^3 +x^2)(x^2 -x + 1) - x^2(x^2 -x +1) -x(x^2 -x +1) -(x^2 -x + 1) = $

$x^4(x^2 -x +1) + x^3(x^2 -x + 1) + x^2(x^2-x +1) -x^4 +x^3 -x^2 -x^3 +x^2 -x -x^2 +x -1 =$

$x^6 -x^5 + x^4 + x^5 -x^4 +x^3 +x^4 -x^3 + x^2 -x^4 -x^2-1=$

$x^6 -1=$

In short, it doesn't matter how you note it. Just note it.

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