How to see $x^6-1=(x^2−1)(x^2+x+1)(x^2−x+1)$?

Question

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?

Answer 1

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

Answer 2

Also, by using your idea we obtain: $$x^6-1=(x^2-1)(x^4+x^2+1)=(x^2-1)(x^4+2x^2+1-x^2)=$$ $$=(x^2-1)((x^2+1)^2-x^2)=(x^2-1)(x^2-x+1)(x^2+x+1).$$

Answer 3

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