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The number of solutions of the equations $x+y+z=1 ; x^2+y^2+z^2=1 ; x^3+y^3+z^3=1$ is
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The correct answer is:
$3$
$x+y+z=1$
$\begin{aligned} & x^2+y^2+z^2=1 \\ & x^3+y^3+z^3=1\end{aligned}$
It is evident from all 3 that equation satisfies only if 2 of them are zero and third is 1.
There are 3 solution sets of $(x, y, z)$
$(0,0,1),(0,1,0)$ and $(1,0,0)$
$\begin{aligned} & x^2+y^2+z^2=1 \\ & x^3+y^3+z^3=1\end{aligned}$
It is evident from all 3 that equation satisfies only if 2 of them are zero and third is 1.
There are 3 solution sets of $(x, y, z)$
$(0,0,1),(0,1,0)$ and $(1,0,0)$
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