The number of solutions of the equations x+y+ | vedclass

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(B) Given equations are:$1) x+y+z=1$$2) x^2+y^2+z^2=1$$3) x^3+y^3+z^3=1$From $(1)$,we have $z = 1 - x - y$. Substituting this into $(2)$:$x^2 + y^2 +

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$3$

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(B) Given equations are:$1) x+y+z=1$$2) x^2+y^2+z^2=1$$3) x^3+y^3+z^3=1$From $(1)$,we have $z = 1 - x - y$. Substituting this into $(2)$:$x^2 + y^2 + (1 - x - y)^2 = 1$$x^2 + y^2 + 1 + x^2 + y^2 - 2x - 2y + 2xy = 1$$2x^2 + 2y^2 + 2xy - 2x - 2y = 0$$x^2 + y^2 + xy - x - y = 0$Multiplying by $2$: $2x^2 + 2y^2 + 2xy - 2x - 2y = 0$,which can be written as $(x+y)^2 + x^2 + y^2 - 2x - 2y = 0$.Alternatively,note that if two variables are $0$,the third must be $1$. For example,if $x=1, y=0, z=0$,then $1+0+0=1$,$1^2+0^2+0^2=1$,and $1^3+0^3+0^3=1$. These satisfy all equations.Testing permutations of $(1, 0, 0)$,we get $(1, 0, 0)$,$(0, 1, 0)$,and $(0, 0, 1)$.Thus,there are $3$ solutions.

The number of solutions of the equations $x+y+z=1$,$x^2+y^2+z^2=1$,and $x^3+y^3+z^3=1$ is:

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