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Title says it all, I just want to know how to factor $(x+y)^4+x^4+y^4$. I only know that it's possible to factor, but got no idea how to do it. If it were a single-variable polynomial I could try to find rational roots or something, but I'm lost with this one.

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  • $\begingroup$ $2x^4 + 4x^3y + 6x^2y^2 + 4^xy^3 + 2y^4$ can be split up as $(2x^4 + 2x^3y + 2x^2y^2) + (2x^3y + 2x^2y^2 + 2xy^3) + (2x^2y^2 + 2xy^3 + 2y^4)$, or $2(x^2 + xy + y^2)^2$. $\endgroup$ – Chris Mar 7 '17 at 20:40
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Beyond what I have posted, it cannot be factored in the real field. If you want to factor it in the complex field, you have to learn how to solve a quartic equation because you essentially need to know that roots of $(t+1)^4 + t^4 + 1 = 2t^4 + 4t^3 + 6t^2 + 4t + 2 = 0$, it can be simplified to $t^4 + 2t^3 + 3t^2 + 2t + 1 = 0$.Lucky enough, this is equal to $(t^2 + t + 1)^2$

Thus $x^4 + y^4 + (x + y)^4 = (x^2 + xy + y^2)^2$. No further factorization is available in the real field...

Best wishes

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  • $\begingroup$ Are you sure about the last line? The coefficient of $x^4$ on the LHS is 2, but the coefficient of $x^4$ on the RHS is 1. Moreover, plugging in $x=1$, $y=0$, we get that LHS = 2, but RHS = 1. What I'm trying to say is, there is clearly something wrong with your last line... [Figured it out - you are missing a '2' in front of the brackets on the RHS of the last equality - I am not being allowed to edit your answer for some reason] $\endgroup$ – John Don Mar 7 '17 at 21:13
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Note that $$(x+y)^4 + x^4 + y^4 = y^4 ((x/y+1)^4 + (x/y)^4 + 1)$$ and see if you can factor $((t+1)^4 + t^4 + 1$. There is a quadratic factor.

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  • $\begingroup$ Just a quick question, did you do this manipulation just so you can have it in one variable? When factoring, is it always best to try to reduce it to one variable? $\endgroup$ – Ovi Mar 7 '17 at 20:43
  • $\begingroup$ This answer seems the most informative and straightforward for the OP - it gives a good general method for factoring polynomials in two variables by reducing it to something in one variable. $\endgroup$ – John Don Mar 7 '17 at 21:19
  • $\begingroup$ This works when a two-variable polynomial is homogeneous. In general, things are more complicated. $\endgroup$ – Robert Israel Mar 8 '17 at 0:56
  • $\begingroup$ @RobertIsrael Ah yes, I didn't mean to be quite so general, but yes, it is a nice reliable method of factoring homogeneous two variable polynomials, which is accessible to those who have not come across many polynomials in more than one variable before. $\endgroup$ – John Don Mar 8 '17 at 19:52
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I know that using complex algebra, we can factor $x^4 + y^4 = (x^2 + y^2 + \sqrt{2}xy)(x^2 + y^2 - \sqrt{2}xy)$. I have no idea how to proceed forward... Good luck

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I am sorry, there is a slight mistake, I forgot factoring 2. Thus $x^4 + y^4 + (x+y)^4 = 2(x^2 + xy + y^2)^2$

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This is a symmetric polynomial in $x$ and $y$, hence it can be expressed, by Newton's theorem, as a polynomial in $s=x+y$ and $p=xy$.

Indeed $x^2+y^2=(x+y)^2-2xy=s^2-2p$, whence $$x^4+y^4=(x^2+y^2)^2-2x^2y^2=(s^2-2p)^2-2p^2=s^4-4ps^2+2p^2.$$ Therefore $$(x+y)^4+x^4+y^4=2s^4-4ps^2+2p^2=2(s^2-p)^2=2(x^2+xy+y^2)^2.$$

$x2+xy+y^2$ is irreducible in $\mathbf R[x]$, but fractors in $\mathbf C[x]$ as $$(x-jy)(x-j^y),\quad\text{where } j\; \text{and }j^2 \;\text{are the non-real cubic roots of unity.}$$

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$$(x+y)^4+x^4+y^4=(x^2+y^2+2xy)^2+(x^2+y^2)^2-2(xy)^2=\\ [(x^2+y^2+2xy)^2-(xy)^2]+[(x^2+y^2)^2-(xy)^2]=\\ (x^2+y^2+xy)(x^2+y^2+3xy)+(x^2+y^2+xy)(x^2+y^2-xy)=\\ (x^2+y^2+xy)(2x^2+2y^2+2xy)=2(x^2+y^2+xy)^2$$

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$$x^4 + y^4 + (x + y)^4 =2 x^4 + 4 x^3 y + 6 x^2 y^2 + 4 x y^3 + 2 y^4$$ $$=x (x (x (2 x + 4 y) + 6 y^2) + 4 y^3) + 2 y^4 =2 x^4 + y (4 x^3 + y (6 x^2 + y (4 x + 2 y)))$$ $$=4 y^2 (x^2 + x y) + 2 (x^2 + x y)^2 + 2 y^4 =2 (x^4 + 2 x^3 y + 3 x^2 y^2 + 2 x y^3 + y^4)$$ $$=2 (x^2 + x y + y^2)^2$$

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