Does a negative exponent distribute?

Question

This should be simple but I am confused.

$$(a^{-1}+b^{-1})^{-1}$$

How is the negative exponent distributed to the terms inside the parentheses? If they were being multiplied it would be simple but they are being added and I'm not sure how to proceed.

Answer 1

$$a^{-1}=\frac1a $$ no matter what a is. a could be an expression or a number, it doesn't matter; the -1 exponent just puts it as a denominator in a fraction with numerator 1.

Thus the answer is $$\frac1{\frac1a +\frac1b}=\frac1{\frac{a+b}{ab}}=\frac{ab}{a+b}$$

Answer 2

$$(a^{-1}+b^{-1})^{-1} = \frac 1 {\frac 1 a + \frac 1 b}= \frac {ab} {a+b}.$$

Answer 3

Exponentiation of real (or complex) numbers never distributes over addition of real (or complex) numbers.

For positive natural numbers, this is because:

$$(a+b)^n=\sum_{k=0}^n\binom{n}{k}a^{n-k}b^{k}$$

This is called the binomial theorem and you can read about it―and its extension to non-positive and non-natural values―at the linked provided herebefore.