How do you simplify ${(\dfrac{{125}}{{64}})^{ - \dfrac{2}{3}}}$ ?

Hint: To solve the given expression, first we should know that the given expression belongs to the property of exponents. So, we will solve this question by using the property of exponent of both division and multiplication.

Complete step-by-step solution:
The given expression belongs to the property of exponents i.e. ${(\dfrac{{125}}{{64}})^{ - \dfrac{2}{3}}}$ .
A property of exponents states that:
${(\dfrac{a}{b})^{ - m}} = {(\dfrac{b}{a})^m}$
Hence,
$
  {(\dfrac{{125}}{{64}})^{ - \dfrac{2}{3}}} \\
   = {(\dfrac{{64}}{{125}})^{\dfrac{2}{3}}} \\
   = {(\dfrac{{64}}{{125}})^{(\dfrac{1}{3}).2}} \\
 $
We also know that ${a^{m.n}} = {({a^m})^n}$
$ = {\{ {(\dfrac{{64}}{{125}})^{\dfrac{1}{3}}}\} ^2}$
Now, again we know that:
\[{a^{\dfrac{1}{m}}} = \sqrt[m]{a}\]
$
   = {(\sqrt[3]{{\dfrac{{64}}{{125}}}})^2} \\
   = {(\sqrt[3]{{\dfrac{{{4^3}}}{{{5^3}}}}})^2} \\
   = {(\sqrt[3]{{{{(\dfrac{4}{5})}^3}}})^2} \\
   = {(\dfrac{4}{5})^2} \\
   = \dfrac{{{4^2}}}{{{5^2}}} \\
   = \dfrac{{16}}{{25}} \\
 $

Hence, the simplified form of the given expression is \[\dfrac{{16}}{{25}}\].

Note: As discussed earlier, there are majorly six laws or rules defined for exponents. Below, all the laws are represented:
(1.) ${a^m}\times {a^n} = {a^{m + n}}$
(2.) ${({a^m})^n} = {a^{m.n}}$
(3.) ${(a\times b)^n} = {a^n}.{b^n}$
(4.) ${(\dfrac{a}{b})^n} = \dfrac{{{a^n}}}{{{b^n}}}$
(5.) \[\dfrac{{{a^m}}}{{{a^n}}} = {a^{m - n}}\]
(6.) $\dfrac{{{a^m}}}{{{a^n}}} = \dfrac{1}{{{a^{m - n}}}}$