If the mean of x, y and z is A and xy is additive inverse of (x + y)z, then…

For any three real numbers x, y, z, the algebraic identity (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) connects their sum, the sum of their squares, and the sum of their pairwise products - this is the key identity used to move from information about the sum to information about the sum of squares.

1. Mean of x, y, z is A, so x + y + z = 3A.

2. "xy is the additive inverse of (x + y)z" means xy = -(x + y)z = -xz - yz, i.e. xy + yz + zx = 0.

3. Apply the identity: (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx).

4. Substitute the known values: (3A)2 = x2 + y2 + z2 + 2(0).

5. This gives x2 + y2 + z2 = 9A2.

6. The mean of x2, y2, z2 is (x2 + y2 + z2) / 3 = 9A2 / 3 = 3A2.

Check with concrete numbers: for A = 1, taking x = 1 + √3, y = 1 - √3, z = 1 gives x + y + z = 3 (so A = 1) and xy + yz + zx = (1 - 3) + (1 - √3) + (1 + √3) = -2 + 2 = 0, satisfying both conditions. Then x2 + y2 + z2 = (4 + 2√3) + (4 - 2√3) + 1 = 9, and the mean of the squares is 9/3 = 3 = 3 × 12, matching 3A2.

So the mean of x2, y2, z2 is 3A2.