If x + y + z = 3 and xy + yz + zx = 3, find x³ + y³ + z³.
3
Using \(x^2+y^2+z^2=(x+y+z)^2-2(xy+yz+zx) = 9-6 = 3\).
So \(x^2+y^2+z^2-(xy+yz+zx) = 3-3 = 0\), which equals \(\frac{1}{2}\left[(x-y)^2+(y-z)^2+(z-x)^2\right]\), forcing \(x=y=z\).
Since \(x+y+z=3\) and \(x=y=z\), each variable equals 1.
Therefore, \(x^3+y^3+z^3 = 1^3+1^3+1^3 = 3\).
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