Fermat's Last Theorem for Negative $n$

Let's suppose that $a^{-n}+b^{-n} = c^{-n}$ for some positive integers $a,b,c,n$.

Then, multiply both sides by $a^nb^nc^n$ to get $b^nc^n + a^nc^n = a^nb^n$, i.e. $(bc)^n+(ac)^n=(ab)^n$.

If $n \ge 3$, then this contradicts Fermat's Last Theorem. Hence, there are no solutions for $n \ge 3$.

For $n = 1$, we have several solutions, one of which is $3^{-1}+6^{-1} = 2^{-1}$.

For $n = 2$, we have several solutions, one of which is $15^{-2}+20^{-2} = 12^{-2}$.

Fermat's theorem actually asks the above question for positive integers greater than two; so are you asking for negative $n$ less than negative two?

Anyways, to answer your question consider the problem for $n = -1$. We have $$ \frac{1}{a} + \frac{1}{b} = \frac{1}{c} = \frac{b}{ab} + \frac{a}{ab} = \frac{a+b}{ab}. $$ This means that if we have $(a+b) \vert ab$ then we can simply define $c=\frac{ab}{a+b}$. There exist many formulas for finding two distinct integers such that their sum divides their product (see link: Necessary and sufficient conditions for the sum of two numbers to divide their product).

Ex: let $a=10$, $b=15$. We have $$ \frac{1}{10}+\frac{1}{15} = \frac{1}{6}. $$