Title says it all, I simply want to know exactly how to element \$(x+y)^4+x^4+y^4\$. Ns only understand that it"s feasible to factor, yet got no idea how to perform it. If it to be a single-variable polynomial i could shot to uncover rational roots or something, but I"m lost with this one.

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\$, however factors in \$altoalsimce.orgbf C\$ as\$\$(x-jy)(x-j^2y)quad extwhere j;  extand j^2 ; extare the non-real cubic roots of unity.\$\$  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...

Best desire

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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