If x + y + z = 0 and x2 + y2 + z2 = 2, find x4 + y4 + z4.
2
When \(x + y + z = 0\), squaring gives \(x^2+y^2+z^2 + 2(xy+yz+zx) = 0\), so \(xy+yz+zx = -\frac{x^2+y^2+z^2}{2} = -1\).
Now, \((x^2+y^2+z^2)^2 = x^4+y^4+z^4 + 2(x^2y^2+y^2z^2+z^2x^2)\), and since \(x^2y^2+y^2z^2+z^2x^2 = (xy+yz+zx)^2 - 2xyz(x+y+z) = (-1)^2 - 0 = 1\).
So \(2^2 = x^4+y^4+z^4 + 2(1)\), giving \(4 = x^4+y^4+z^4 + 2\).
So \(x^4+y^4+z^4 = 2\).
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