Question 15149
In order to answer {{{64^(-5/6)}}} you have to understand a few properties of exponents.

{{{a^(-b) = 1/a^b}}}
{{{a^b*a^d = a^(b+d)}}}
{{{(a^b)^c=a^(bc)}}}

Those rules also apply to fracial exponents. There are two parts to a fractional exponent: the numberator of the exponent, which is the power index, and the denominator of the exponent with the root index. Therefore, {{{a^(1/2) = sqrt(a)}}} and so on.

To your problem, 
{{{64^(-5/6)}}}

First handle the negative exponent. As shown in one of the properties I illustrated, a negative exponent just means "flip it". 
{{{(1/64)^(5/6)}}}
Then, you can do the numerator first or the denominator. {{{64^5}}} is going to be pretty big so lets do the denominator first.
{{{((1/64)^(1/6))^5}}}

The 6th root of 64, meaning then number that satisfied {{{x^6=64}}} is 2. Therefore you have: 
{{{(1/2)^5}}}

2^5 is 32, so the final answer is:

{{{1/32}}}