Calculating xyz from Given Equations

[aznluster]

Given:

$x+y+z=1$ $x^{2}+y^{2}+z^{2}=2$$x^{4}+y^{4}+z^{4}=4$

Find:$x^{3}+y^{3}+z^{3}$ Attempt at Solving:Note: $$x^{3}+y^{3}+z^{3}-3xyz=(x+y+z)(x^{2}+y^{2}+z^{2}-xy-xz-yz)$$$(x+y+z)^{2}=x^{2}+y^{2}+z^{2}+2xy+2xz+2zy=1$$xy+xz+zy=-1/2$ Plugging in the unknown, we get:$x^{3}+y^{3}+z^{3}-3xyz=(1)(2-(-1/2))$ Now I need to find $xyz$.

 

[phyzguy]

Using:
$$(x^2+y^2+z^2)^2=x^4+y^4+z^4+ 2 (x^2 y^2+x^2 z^2+y^2<br /> z^2)$$
$$(x+y+z)^4=x^4+y^4+z^4+4 (x y+x z+y z) (x^2+y^2+z^2)+6<br /> (x^2 y^2+x^2 z^2+y^2 z^2)+8 x y z (x+y+z)$$

You can solve the first for x^2 y^2+x^2 z^2+y^2 z^2, and then the second for xyz.

 

[aznluster]

Thank you. :)