Express ${\left(\frac{64}{729}\right)}^{\left(-\frac{1}{6}\right)}$  in the simplest form.

Given: ${\left(\frac{64}{729}\right)}^{\left(-\frac{1}{6}\right)}$ 

To simplify the expression, first convert the negative exponent to positive.
${\left(\frac{64}{729}\right)}^{\left(-\frac{1}{6}\right) }= {\left(\frac{729}{64}\right)}^{\left(\frac{1}{6}\right)}$ 

Now do prime factorization of both the numerator and the denominator then use the power of quotient property and the property of a power raised to another power.
$\begin{array}{c}{\left(\frac{729}{64}\right)}^{\left(\frac{1}{6}\right)}={\left(\frac{3\times 3\times 3\times 3\times 3\times 3}{2\times 2\times 2\times 2\times 2\times 2}\right)}^{\left(\frac{1}{6}\right)}\\ ={\left(\frac{3^6}{2^6}\right)}^{\left(\frac{1}{6}\right)}\\ = {\left(\frac{3}{2}\right)}^{\left(6\times \frac{1}{6}\right)}\end{array}$ 
                            $= \frac{3}{2}$

So, the simplest form of ${\left(\frac{64}{729}\right)}^{\left(-\frac{1}{6}\right)}$  is obtained as  $\frac{\mathbf{3}}{\mathbf{2}}$.