Дам 50 балловx+y+z=1x²+y²+z²=2x³+y³+z³=3x⁴+y⁴+z⁴=?

Решите задачу:

$(x + y + z) ^{2} = 1$
$x^{2} + y^{2} + z^{2} +2 (xy + xz + yz) = 1$
$2 +2 (xy + xz + yz) = 1$
$(xy + xz + yz) = -\frac{1}{2}$
$(xy + xz + yz)^{2} = \frac{1}{4}$
$(xy)^{2} + (xz) ^{2} + (yz)^{2} +2xyz(x + y + z)= \frac{1}{4}$
$(xy)^{2} + (xz) ^{2} + (yz)^{2} +2xyz= \frac{1}{4}$

$x^{3}+y^{3}+z^{3}-3xyz=(x + y + z)(x^{2}+y^{2} +z^{2}-xy-xz-yz)=$ $1*(2+\frac{1}{2})=2\frac{1}{2}$
$3xyz=3-2\frac{1}{2}=\frac{1}{2}$
$xyz=\frac{1}{6}$
$(xy)^{2} + (xz) ^{2} + (yz)^{2} +2*\frac{1}{6}= \frac{1}{4}$
$(xy)^{2} + (xz) ^{2} + (yz)^{2}=- \frac{1}{12}$

$(x^{2} + y^{2} + z^{2})^{2} = 4$
$x^4+y^4+z^4+2((xy)^{2} + (xz) ^{2} + (yz)^{2}) =4$
$x^4+y^4+z^4=4-2*(- \frac{1}{12})=4\frac{1}{6}$