Question

Prove that `(a^x/a^y)^(x + y) (a^y/a^z)^(y + z) (a^z/a^x)^(z + x) = 1`.

Answer 1

Given:

`(a^x/a^y)^(x + y) (a^y/a^z)^(y + z) (a^z/a^x)^(z + x)`

To Prove:

`(a^x/a^y)^(x + y) (a^y/a^z)^(y + z) (a^z/a^x)^(z + x) = 1`

Proof (Step-wise):

1. Rewrite each fraction inside the parentheses using properties of exponents:

`a^x/a^y = a^(x - y)`

`a^y/a^x = a^(y - z)`

`a^z/a^x = a^(z - x)`

2. Substitute these back into the expression:

`(a^(x - y))^(x + y) (a^(y - z))^(y + z) (a^(z - x))^(z + x)`

3. Use the power of a power rule (a^m)ⁿ = a^mn:

`a^((x - y)(x + y)) xx a^((y - z)(y + z)) xx a^((z - x)(z + x))`

4. Evaluate products inside the exponents using the identity (m – n)(m + n) = m² – n²:

`a^(x^2 - y^2) xx a^(y^2 - z^2) xx a^(z^2 - x^2)`

5. Combine powers by adding exponents of the same base:

`a^((x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2)) = a^0`

`a^((x^2 - y^2) + (y^2 - z^2) + (z^2 - x^2)) = 1`

`(a^x/a^y)^(x + y) (a^y/a^z)^(y + z) (a^z/a^x)^(z + x) = 1`

Hence, the expression is proven to be equal to 1.