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Question Number 46882 by maxmathsup by imad last updated on 02/Nov/18

factorize inside C[x] x^2 +y^2 +z^2

factorizeinsideC[x]x2+y2+z2

Commented by maxmathsup by imad last updated on 02/Nov/18

we have (x+y+z)^(2 ) =x^(2 ) +y^2 +z^2 +2(xy +yz +zx) ⇒ x^2 +y^2 +z^2 =(x+y+z)^2 −((√(2(xy +yz +zx)))^2 let 2(xy+ yz +zx)=r e^(iθ) ⇒ x^2 +y^2 +z^2 =(x+y+z)^2 −((√r)e^(i(θ/2)) )^2 =(x+y+z −(√r)e^(i(θ/2)) )(x+y+z +r e^(i(θ/2)) ) .

wehave(x+y+z)2=x2+y2+z2+2(xy+yz+zx)⇒x2+y2+z2=(x+y+z)2−(2(xy+yz+zx)2let2(xy+yz+zx)=reiθ⇒x2+y2+z2=(x+y+z)2−(reiθ2)2=(x+y+z−reiθ2)(x+y+z+reiθ2).

Commented by maxmathsup by imad last updated on 02/Nov/18

another method we have x^2 +y^2 +z^2 =x^2 −(i(√(y^(2 ) +z^2 )))^2 =(x−i(√(y^2 +z^2 )))(x+i(√(y^2 +z^2 ))) or x^2 +y^2 +z^2 =x^2 +y^2 −(iz)^2 =((√(x^2 +y^2 ))−iz)((√(x^2 +y^2 ))+iz)

anothermethodwehavex2+y2+z2=x2−(iy2+z2)2=(x−iy2+z2)(x+iy2+z2)orx2+y2+z2=x2+y2−(iz)2=(x2+y2−iz)(x2+y2+iz)

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