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I am working on factoring $a^4+b^4$ and I have found two different solutions to this.

First, I have factored it to $$a^4+b^4=(a+b)[a^3-a^2b+ab^2+b^3]-2ab^3$$

But then I also found that the equation is factorable to $$a^4+b^4=(a^2+b^2-\sqrt{2}ab)(a^2+b^2+\sqrt{2}ab)$$

First of all, why are there two ways of factoring this equation, and why are the solutions not the same?

Additionally, is it possible to convert either solution so it is the same as the other?

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  • $\begingroup$ Because the first one is wrong. $\endgroup$ – quasi May 29 '17 at 14:52
  • $\begingroup$ One of your factorizations is not correct. There is an easy way of checking which one, as the "solutions" it gives you are actually not solutions. $\endgroup$ – Andrés E. Caicedo May 29 '17 at 14:53
  • $\begingroup$ Alpha shows the first is correct. $\endgroup$ – Ross Millikan May 29 '17 at 15:07
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    $\begingroup$ Yeah I don't get why everyone is saying either of those is wrong. It's wrong to call the first one a factorization but the equation itself isn't wrong. The equation is true. We need to be more explicit about what we're calling "wrong" here. $\endgroup$ – tilper May 29 '17 at 15:12
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Contrary to what's being said in the comments, neither of those expressions is wrong because both yield $a^4 + b^4$ when expanded. It is, however, incorrect to call this a factorization:

$$a^4+b^4=(a+b)[a^3-a^2b+ab^2+b^3]-2ab^3$$

When you factor an expression, say $F(x)$, you break it down into two or more factors, such as $F(x) = G(x)H(x)K(x)$. That is not what you did here. You broke it down into two factors plus a remainder of $-2ab^3$. That is not factoring.

To address the question more generally, since $\Bbb R$ (or $\Bbb C$, or $\Bbb Q$, whichever you're working with) is a field, then $\Bbb R[x]$ is a unique factorization domain (among other things, but those other things are irrelevant here). This means that every expression in $\Bbb R[x]$ has one and only one factorization (up to ordering of the factors, for example $x(x-1)$ and $(x-1)x$ are considered the same factorization of $x^2-x$). An expression may have two factorizations that look different but actually aren't.

For a simple example over $\Bbb C[x]$, we could say $x^4 - 1 = (x^2 + 1)(x^2 - 1)$ and we could say $x^4 - 1 = (x^2 - x + i(1-x))(x^2 + x + i(1+x))$. These look very different but they're really the same. How can we be sure they're the same? Break it into linear factors over $\Bbb C$: $$ x^4 - 1 = (x-1)(x+1)(x+i)(x-i)$$

$(x-1)(x+1) = x^2-1$ and $(x+i)(x-i) = x^2+1$. That's how we can get the first factorization.

Alternatively, $(x-1)(x-i) = x^2 - x + i(1-x)$ and $(x+1)(x+i) = x^2 + x + i(1+x)$. That's how we can get the second factorization.

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    $\begingroup$ Ah, thanks for the explanation. It was actually something I was asked by a 12-year-old girl I tutor a bit. Was starting to feel bad about not being able to answer her so I finally asked on here. $\endgroup$ – Anders T May 30 '17 at 18:37
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A factorization is always a pure product, e.g.,

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

or

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

So the problem is that $(a+b)(a^3-a^2b+ab^2+b^3)-2ab^3$, is not a pure product. Instead, it's the sum of the product $(a+b)(a^3-a^2b+ab^2+b^3)$ and $-2ab^3$.

I.e., what tilper just said!

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