Question

x+y+z=0 x²+y²+z²=1 x四次方+y四次方+z四次方=？ 答得好加分

这个题很难，希望大家帮个忙

Answer 1

(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)=0，因为x^2+y^2+z^2=1，所以xy+yz+zx=-1/2；
(xy+yz+zx)^2=1/4=(xy)^2+(yz)^2+(zx)^2+2xyz(x+y+z)，所以(xy)^2+(yz)^2+(zx)^2=1/4；
(x^2+y^2+z^2)^2=1=x^4+y^4+z^4+2[(xy)^2+(yz)^2+(zx)^2]，所以x^4+y^4+z^4=1/2。

Answer 2

（x+y+z）^2=[(x+y)+z]^2
=(x^2+2xy+y^2)+z^2+2zx+2zy
=x^2+y^2+z^2+2xy+2xz+2yz
=x^2+y^2+z^2+2(xy+xz+yz)=0
x+y+z=0
xy + xz+yz= -1/2

(xy+xz+yz)^2
=x^2y^2+x^2z^2+y^2z^2+2xzy^2+2yzx^2+2xyz^2
=x^2y^2+x^2z^2+y^2z^2 +2xyz(x+y+z)
=1/4
x^2y^2+x^2z^2+y^2z^2=1/4
(x^2+y^2+z^2)^2
=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2=1
x^4+y^4+z^4= 1/2