Since the square of any real number is always greater than or equal to zero, the only way the sum of three squares can be zero is if each square is zero. Thus, x = y = z.
Now, using the second equation xyz = 1 and x = y = z, we get:
x^3 = 1 x = 1
So, the solution to the given system of equations is x = y = z = 1.
The given equations are a system of non-linear equations. By rearranging the first equation we get:
x^2 + y^2 + z^2 - x - y - z = 0
By adding and subtracting 2(xy + yz + zx) to the equation, we get:
x^2 - 2xy + y^2 + y^2 - 2yz + z^2 + z^2 - 2zx + x^2 = 0
(x - y)^2 + (y - z)^2 + (z - x)^2 = 0
Since the square of any real number is always greater than or equal to zero, the only way the sum of three squares can be zero is if each square is zero. Thus, x = y = z.
Now, using the second equation xyz = 1 and x = y = z, we get:
x^3 = 1
x = 1
So, the solution to the given system of equations is x = y = z = 1.