Solve the following equations : $x+y+z=1,\ x^{2}+y^{2}+z^{2}=1,\ x^{3}+y^{3}+z^{3}=1.$
- $(1,0,0),(0,2,0),(0,0,1)$.
- $(1,0,0),(0,1,0),(0,0,1)$.
- $(1,0,0),(0,1,0),(0,0,4)$.
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None of these
Given x+y+z=1, x^2+y^2+z^2=1, x^3+y^3+z^3=1. If x, y, z are 1, 0, 0 in any order, the equations are satisfied: 1+0+0=1, 1^2+0^2+0^2=1, 1^3+0^3+0^3=1. These are the only real solutions.
Using the algebraic identity, x squared plus y squared plus z squared equals the square of (x plus y plus z) minus 2 times (xy plus yz plus zx). Substituting the given values where both sums equal 1 results in 1 equaling 1 minus 2 times (xy plus yz plus zx), meaning xy plus yz plus zx equals 0. Another standard identity states that x cubed plus y cubed plus z cubed minus 3xyz equals the quantity (x plus y plus z) times (x squared plus y squared plus z squared minus xy minus yz minus zx), which simplifies to 1 minus 3xyz equals 1, giving xyz equals 0. Since xyz equals 0, at least one of the variables is zero, which forces the other two variables to multiply to zero from xy plus yz plus zx equals 0, leaving exactly one variable equal to 1 and the other two equal to 0.