Fill in the blank to make the following statement true:
$(64)^{-\frac{1}{6}} = \ldots \ldots$

(1/2) To solve $(64)^{-\frac{1}{6}}$,we first express $64$ as a power of $2$.
We know that $64 = 2^6$.
Substituting this into the expression,we get:
$(2^6)^{-\frac{1}{6}}$
Using the exponent rule $(a^m)^n = a^{m \times n}$,we multiply the exponents:
$2^{6 \times (-\frac{1}{6})} = 2^{-1}$
Since $a^{-n} = \frac{1}{a^n}$,we have:
$2^{-1} = \frac{1}{2^1} = \frac{1}{2}$