If x + y + z = 4, x2 + y2 + z2 = 6 and x-1 + y-1 + z-1 = 1, where x, y, z are…

Concept: For any three numbers x, y, z, the identity (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx) relates their sum, sum of squares, and sum of pairwise products. Also, when x, y, z are all non-zero, the sum of reciprocals satisfies 1/x + 1/y + 1/z = (xy + yz + zx) / (xyz).

Application:

1. Given x + y + z = 4, x2 + y2 + z2 = 6, and 1/x + 1/y + 1/z = 1.

2. Square the first equation: (x + y + z)2 = 42 = 16.

3. Apply the identity: 16 = x2 + y2 + z2 + 2(xy + yz + zx) = 6 + 2(xy + yz + zx).

4. Solve for the pairwise-product sum: 2(xy + yz + zx) = 16 − 6 = 10, so xy + yz + zx = 5.

5. Use the reciprocal identity: 1/x + 1/y + 1/z = (xy + yz + zx) / (xyz), so 1 = 5 / (xyz).

6. Solve for the product: xyz = 5.

Cross-check: Re-deriving the pairwise-product sum from x2 + y2 + z2 = (x + y + z)2 − 2(xy + yz + zx) gives 6 = 16 − 2(xy + yz + zx), i.e., xy + yz + zx = 5 — the same value obtained above, confirming the step. Substituting this into 1/x + 1/y + 1/z = (xy + yz + zx)/xyz with the given reciprocal sum of 1 leaves xyz = 5 as the only consistent value.

Note: This derivation uses only the algebraic identities above and holds whether x, y, z are real or complex; here they are precisely the three roots (one real, two complex conjugates) of t3 − 4t2 + 5t − 5 = 0, which indeed satisfy all three given equations, confirming the system is consistent.

Therefore, xyz = 5.