Originally Posted by
simplependulum Suppose $\displaystyle x,y,z ~~$ are the roots of a cubic equatiox^3n
we have $\displaystyle x+y+z = 1$ so the equation is $\displaystyle
u^3 - u^2 + bu + c = 0 $
At the same time ,
$\displaystyle x^2 + y^2 + z^2 = (x+y+z)^2 - 2(xy+yz+zx) = 1$ so
$\displaystyle 1 - 2(xy+yz+zx) = 1 \implies xy+yz+zx = 0$
The equation :$\displaystyle u^3 - u^2 + c = 0 $
And
$\displaystyle x^3 = x^3 - x^2 + c + ( x^2 - x + \frac{c}{x}) - c + x - \frac{c}{x} = 0+0 - c + x - \frac{c}{x} = x - c - \frac{c}{x}$
So $\displaystyle x^3 + y^3 + z^3 = x+y+z - c-c-c - c (\frac{1}{x} + \frac{1}{y} + \frac{1}{z}) = 1 - 3c - c [ \frac{ xy+yz+zx}{xyz}] = 1-3c $
It is given that $\displaystyle x^3 + y^3 + z^3 = 1$
Therefore , $\displaystyle 1 = 1-3c \implies c = 0$
The equation is u^3 - u^2 = 0
u = 0 , 0, 1
Therefore , (x,y,z) = (0,0,1),(0,1,0),(1,0,0)
If the values of the right hand sides are not 1,the above method can be applied to solve it , It is not just used to prove the set of the solution is (0,0,1) in this case .