You are watching: Factor x^4-y^4

Note the $$(x+y)^4 + x^4 + y^4 = y^4 ((x/y+1)^4 + (x/y)^4 + 1)$$and see if friend can element $((t+1)^4 + t^4 + 1$. Over there is a quadratic factor.

This is a symmetric polynomial in $x$ and $y$, for this reason it can be expressed, by

*Newton"s theorem*, as a polynomial in $s=x+y$ and also $p=xy$.

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

$x^2+xy+y^2$ is irreducible in $altoalsimce.orgbf R

Beyond what I have actually posted, it can not be factored in the actual field. If you want to aspect it in the complicated field, you have to learn exactly how to settle a quartic equation since you essentially need to know that root of $(t+1)^4 + t^4 + 1 = 2t^4 + 4t^3 + 6t^2 + 4t + 2 = 0$, it deserve to be simplified to $t^4 + 2t^3 + 3t^2 + 2t + 1 = 0$.Lucky enough, this is same to $(t^2 + t + 1)^2$

Thus $x^4 + y^4 + (x + y)^4 = (x^2 + xy + y^2)^2$.No additional factorization is easily accessible in the genuine field...

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

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

$$(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$$

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

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