Title says it all, I just want to know how khổng lồ factor $(x+y)^4+x^4+y^4$. I only know that it"s possible to factor, but got no idea how khổng lồ vị 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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Beyond what I have sầu posted, it cannot be factored in the real field. If you want to lớn factor it in the complex field, you have sầu lớn learn how to solve sầu a quartic equation because you essentially need lớn know that roots of $(t+1)^4 + t^4 + 1 = 2t^4 + 4t^3 + 6t^2 + 4t + 2 = 0$, it can be simplified to lớn $t^4 + 2t^3 + 3t^2 + 2t + 1 = 0$.Lucky enough, this is equal khổng lồ $(t^2 + t + 1)^2$

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


Note that $$(x+y)^4 + x^4 + y^4 = y^4 ((x/y+1)^4 + (x/y)^4 + 1)$$và see if you can factor $((t+1)^4 + t^4 + 1$. There is a quadratic factor.


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

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

$x^2+xy+y^2$ is irreducible in $ eftekumsk.combf R$, but factors in $ eftekumsk.combf C$ as$$(x-jy)(x-j^2y)quad extwhere j;  extvà j^2 ; extare the non-real cubic roots of unity.$$


I know that using complex algebra, we can factor $x^4 + y^4 = (x^2 + y^2 + sqrt2xy)(x^2 + y^2 - sqrt2xy)$. I have no idea how khổng lồ proceed forward...Good luck

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


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

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