The value of a + b + c satisfying xyz = 1 (x, y, z ≠ 1), if x(1/a) = y(1/b) =…

Concept:

When several expressions raised to fractional powers all equal the same value, that common value can be treated as a single positive-real base k (x, y, z are taken as positive reals, as is standard for expressions of the form x^(1/a), so k is a positive real number), and each original quantity can be rewritten as a power of k. The stem states x, y, z ≠ 1, so k ≠ 1 as well (k = 1 is exactly the case x = y = z = 1, which is excluded because it would trivially satisfy xyz = 1 for any a, b, c and make a + b + c undetermined). For a positive real k ≠ 1, a power of k equals 1 only when its exponent equals 0 — that is, kn = 1 for positive real k ≠ 1 requires n = 0.

Application:

1. Let x(1/a) = y(1/b) = z(1/c) = k, the common (non-trivial) value of the three fractional powers.

2. Raising each equality to the matching power gives x = ka, y = kb, and z = kc.

3. Multiplying the three expressions: xyz = ka · kb · kc = k(a+b+c).

4. The problem states xyz = 1, so k(a+b+c) = 1.

5. Since k ≠ 1, this forces the exponent a + b + c to equal 0 (the concept step above).

Cross-check:

Pick numbers that fit the pattern, e.g. a = 1, b = 1, c = −2, with k = 2: x = 21 = 2, y = 21 = 2, z = 2(−2) = 1/4, so xyz = 2 × 2 × 1/4 = 1, and a + b + c = 1 + 1 − 2 = 0 — confirming the identity.

Hence a + b + c = 0.