Solution to puzzle 156: Three simultaneous equations

Find all positive real solutions of the simultaneous equations:

x + y² + z³ = 3 (1)
y + z² + x³ = 3 (2)
z + x² + y³ = 3 (3)

Clearly one solution is x = y = z = 1. We shall show that this is the only solution.

From (1) − (2),
x(1 − x²) + y(y − 1) + z²(z − 1) = 0 (4)
Similarly,
y(1 − y²) + z(z − 1) + x²(x − 1) = 0 (5)

Next, from (4) − z · (5),
x(x − 1)(1 + x + xz) = y(y − 1)(1 + z + yz) (6)
Similarly,
y(y − 1)(1 + y + yx) = z(z − 1)(1 + x + zx) (7)

Since x, y, and z are positive, the factors (1 + x + xz), (1 + z + yz), and (1 + y + yx) are all positive.
Hence (x − 1), (y − 1), are (z − 1) are all negative, all zero, or all positive.
That is, x, y, and z are all less than 1, equal to 1, or greater than 1.
We have already accounted for the second case. The other two cases are inconsistent with equations (1), (2) and (3).

Therefore, the only positive real solution of the simultaneous equations is x = y = z = 1.

Source: Polynomials, by Edward J. Barbeau. Problem 8.50.