# factoring $(x^6 - y^6)$: what is going on here?

I apologize if this question already exists, but it was quite difficult to word.

In my math class today, we learned how to factor a difference of two perfect cubes. One of our practice questions was:

factor $x^6 - 64$

almost everyone other than me (including my teacher, as he was rushing) ended up with:

$(x^2 - 4)(x^4 + 4x^2 + 16)$

I pointed out two things:

1:

$(x^6 - 64)$ was also a difference of two perfect squares, and we could factor it into this:

$(x^3 + 8)(x^3 - 8)$

Which could then be factored further into:

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

and 2:

the $(x^2 - 4)$ was also a difference of two perfect squares, and we could factor further:

$(x + 2)(x - 2)(x^4 + 4x^2 + 16)$

Now for my question:

How are $(x + 2)(x^2 - 2x + 4)(x - 2)(x^2 + 2x + 4)$ and $(x + 2)(x - 2)(x^4 + 4x^2 + 16)$ equivalent?

The teacher didn't seem to know off the top of his head, and I can't figure it out after trying for a half an hour. I've heard that you can factor the sum of two perfect squares with imaginary numbers, so maybe i can do something there to help explain this?

edit: fixed instances of "- 16" to "+ 16", and "4x" to "4x^2" (thanks for pointing that out)

• It should read $x^6-64=(x^2-4)(x^4+4x^2+16)$. And $$x^4+4x^2+16=(x^4+8x^2+16)-4x^2=(x^2+4)^2-(2x)^2...$$ Yes, it is very likely that something similar has been handled earlier. – Jyrki Lahtonen Nov 16 '16 at 16:45
• @Arthur, the quartic is actually $x^4+4x^2+16$, not $x^4+4x+16$. I.e., there were two errors, not just one. – Barry Cipra Nov 16 '16 at 18:08
• $(x^2 - 2x + 4)(x^2 + 2x + 4) = x^4 + 4x^2 + 16$ so ... why not? – fleablood Nov 16 '16 at 21:36

## 5 Answers

It's a very common mistake (*) to think that, since $t^2+4t+16$ is irreducible over the reals, also the polynomial obtained with $t=x^2$ is irreducible.

Over the reals, every polynomial of degree $>2$ is reducible, so also $x^4+4x^2+16$ is reducible.

The trick here is to push in a difference of squares: if you add and subtract $4x^2$, you get $$x^4+8x^2+16-4x^2=(x^2+4)^2-(2x)^2=(x^2+4-2x)(x^2+4+2x)$$ so you get the same factorization as with the other method.

Another funny example is $x^4+1$: $$x^4+1=x^4+2x^2+1-2x^2= (x^2+1)^2-(\sqrt{2}\,x)^2= (x^2+1-\sqrt{2}\,x)(x^2+1+\sqrt{2}\,x)$$ which is the full factorization over the reals, since those degree $2$ polynomials have no real roots.

(*) Even among teachers, unfortunately.

If everyone has ended up, as you say, with the factorization $x^6 - 64 = (x^2 - 4)(x^4 + 4x - 16)$, then everybody was wrong: if you take a look at only the free terms (i.e. the terms containing no $x$), you will see that they are $-64$ in the left-hand side, and $(-4)(-16) = +64$ in the right-hand side - which is impossible, therefore this factorization is not correct. The correct one is the one that you have produced. (The fact that your teacher had troubles with this one is a strong signal that you should move to another school - or even move to a country with a functional education system.)

• I had been recreating the equations based on memory, and I believe that everyone had actually written "+ 16". Sorry about that, and thank you for your answer. – Evan Nov 16 '16 at 21:33

There are two errors in the factorization

$$x^6-64=(x^2-4)(x^4+4x-16)$$

The constant term in the quartic should be $+16$ instead of $-16$, and the linear term, $4x$, should be a quadratic term, $4x^2$. That is, the correct factorization is

$$x^6-64=(x^2-4)(x^4+4x^2+16)$$

This now agrees with the factorization $(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)$ since $(x-2)(x+2)=x^2-4$ and

\begin{align} (x^2-2x+4)(x^2+2x+4)&=(x^2+4-2x)(x^2+4+2x)\\ &=(x^2+4)^2-(2x)^2\\ &=x^4+8x^2+16-4x^2\\ &=x^4+4x^2+16 \end{align}

A complete factorization of $x^6 - y^6$ can be performed as follows: \begin{align*} x^6 - y^6 &= (x^3)^2 - (y^3)^2 \\ &= (x^3 - y^3)(x^3 + y^3) \\ &= (x-y)(x^2 + xy + y^2)(x+y)(x^2 - xy + y^2). \end{align*} The quadratic factors are irreducible in the ring of polynomials with rational coefficients $\mathbb Q[x]$ (more informally, they don't factor "nicely"). We can recombine these factors in carefully chosen ways to get expressions that might have resulted from an incomplete factorization; e.g., if we let $a = x^2 + y^2$ and $b = xy$, then \begin{align*} (x^2 + xy + y^2)(x^2 - xy + y^2) &= (a+b)(a-b) \\ &= a^2 - b^2 \\ &= (x^2+y^2)^2 - (xy)^2 \\ &= x^4 + 2x^2 y^2 + y^4 - x^2 y^2 \\ &= x^4 + x^2 y^2 + y^4. \end{align*} And if we combine $(x-y)(x+y)$ we get $x^2 - y^2$; thus we recover the incomplete factorization of $x^6 - y^6$ as a difference of cubes $$(x^2)^3 - (y^2)^3 = (x^2 - y^2)(x^4 + x^2 y^2 + y^4).$$ There are of course other ways to combine these factors, each leading to a different expression, but all are algebraically equivalent.

• The OP cannot perform some polynomial multiplications and you write that "The quadratic factors are irreducible in the ring of polynomials with rational coefficients $\Bbb Q[x]$"? Notice, too, that the OP isn't asking about how to factorize, but rather about why two factorizations are equivalent. – Alex M. Nov 16 '16 at 17:49

"How are $(x+2)(x^2−2x+4)(x−2)(x^2+2x+4)$ and $(x+2)(x−2)(x^4+4x^2+16)$ equivalent?"

Because $(x^2−2x+4)(x^2+2x+4)=x^4+4x^2+16$

===

Nother way of looking at it.

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

$a^2-b^2=(a-b)(a+b)$

So therefore $a^6-b^6=(a^2-b^2)(a^4+a^2b^2+b^4) =(a-b)(a+b) (a^4+a^2b^2+b^4)$

Yet, $a^6-b^6=(a^3-b^3)(a^3+b^3)=(a-b)(a^2+ab+b^2)(a^3+b^3)$

So it must be that $(a+b) (a^4+a^2b^2+b^4)= (a^2+ab+b^2)(a^3+b^3)$ somehow.

So $(a+b)|(a^2+ab+b^2)$ or $(a+b)|(a^3+b^3)$.

$a^2+ab+b^2=a (a+b)+b^2$ which doesn't seem to work.

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

So now we must have $(a+b) (a^4+a^2b^2+b^4)= (a+b)(a^2+ab+b^2)(a^2-ab+b^2)$ and so it must be that:

$a^4+a^2b^2+b^4=(a^2+ab+b^2)(a^2-ab+b^2)$

And if we try to multiply it out, we see that it is so.

I don't blame the students for not seeing this, but the teacher ought to have recognize that it had to happen.

• Now does this mean that (x^4 + 4x^2 + 16) isn't factored completely? if so, then I'm surprised, as my math teacher said that (x+2)(x−2)(x4+4x2+16) was factored completely. – Evan Nov 16 '16 at 21:45
• I might have made a mistake but I don't think so and I'm a bit concerned you teacher didnvt catch this. $(x^2+4 -2×)(x^2+4+2x)=(x^2+4)^2 -4x^2=x^4+8x^2 ×16-4x^2=x^4+4x^2+16$. Yes it factors. – fleablood Nov 16 '16 at 22:24