1) Observe the trivial solution x=y=z=0.

2) Observe that if x, y, z is a solution, then so is x¢ = -x, y¢ = -y, z¢ = -z.

3) If x=y=z is to be a solution, then

(x+y)³=z

Þ

(2x)³=x

If x ¹ 0 [which impiles x=y=z=0] then

x=y=z=1/Ö8 [the observation (2) above covering the negative solution]

4) Without loss of generality (and having recorded the solution (1)) we can assume

x ³ y ³ z > 0. But x < 1 [else (x+y)³=z Þ z > 1, and then that y > 1. But they can't all be > 1]. So

1 > x ³ y ³ z > 0

5) What if z > 1/Ö8? Write z=m/Ö8. Then (x+y)³ ³ (2m/Ö8)³=8m³/8Ö8=m³/ Ö8. The latter is > m/Ö8=z so the first equation is not satisfied. So z < 1/Ö8 [no need for 'less than or equal to' here since equality would imply that they are all equal, which is a solution we already have]

6) Having recorded solutions (1), (3), any reamining solution must have z < 1/Ö8, from (5). Can we have y ³ 1/Ö8? No, because if so, x=m/Ö8, y=b/Ö8 where m and b are both ³ 1. But then, (x+y)³=(m+b)³/8Ö8 ³ 1/Ö8, which is strictly greater than z, so the first equation is not satisfied. So both y and z are < 1/Ö8. The second equation implies x < 1/Ö8 also. So it remains to see whether there is a solution where

1/Ö8 > x ³ y ³ z > 0

The answer is no. Write

x=a/Ö8

y=b/Ö8

z=c/Ö8

Then (y+z)³=1/8Ö8(b+c)³ £ 1/8Ö8 (b+b)³ = 1/Ö8b³ < 1/Ö8b. The latter must be < x, therefore the first equation is not satisfied.

So the only solutions are the ones described in (1), (2), (3).