How do you simplify ${{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}$ ?

Hint: Since, we have to find the simplified value of the given question ${{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}$ , we will follow the following way of solution method as:
$\Rightarrow {{\left( \dfrac{n}{m} \right)}^{\left( -\dfrac{a}{b} \right)}}={{\left( \dfrac{1}{{}^{n}\!\!\diagup\!\!{}_{m}\;} \right)}^{\left( \dfrac{a}{b} \right)}}={{\left( 1\times \dfrac{m}{n} \right)}^{\left( \dfrac{a}{b} \right)}}={{\left( \dfrac{m}{n} \right)}^{\left( \dfrac{a}{b} \right)}}$
Since, the power of the number is also in fraction, so we will use the solution method as:
$\Rightarrow {{\left( \dfrac{m}{n} \right)}^{\left( \dfrac{a}{b} \right)}}={{\left( \dfrac{m}{n} \right)}^{\left( a\times \dfrac{1}{b} \right)}}={{\left[ {{\left( \dfrac{m}{n} \right)}^{\dfrac{1}{b}}} \right]}^{a}}$
Now, we will solve the above term so that we can get the simplified value of the question.

Complete step by step solution:
Since, the question is given as ${{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}$ . Here the both the base term and the power of the number are fraction. Firstly, we will convert the power term from negative to positive as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left( \dfrac{1}{{}^{8}\!\!\diagup\!\!{}_{27}\;} \right)}^{\dfrac{2}{3}}}$
We can write the above equation as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left( 1\div \dfrac{8}{27} \right)}^{\left( \dfrac{2}{3} \right)}}$
If the dividend is in fraction, we can chance the place of nominator and denominator as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left( 1\times \dfrac{27}{8} \right)}^{\left( \dfrac{2}{3} \right)}}$
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left( \dfrac{27}{8} \right)}^{\left( \dfrac{2}{3} \right)}}$
Here, we can expand the above number as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left( \dfrac{27}{8} \right)}^{\left( 2\times \dfrac{1}{3} \right)}}$
Now, we can write the above term as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left[ {{\left( \dfrac{27}{8} \right)}^{\left( \dfrac{1}{3} \right)}} \right]}^{2}}$
Since, $8$ and $27$ can be written as ${{2}^{3}}$ and ${{3}^{3}}$respectively. So, The above equation can be written as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left[ {{\left( \dfrac{{{3}^{3}}}{{{2}^{3}}} \right)}^{\left( \dfrac{1}{3} \right)}} \right]}^{2}}$
Now, the power of the term is in fraction and the fraction is $\dfrac{1}{3}$ that means we will have to find the cube root of the base. The power of the power can be written in the term of multiplication. So, the above term can be written as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left[ \left( \dfrac{{{3}^{3\times \dfrac{1}{3}}}}{{{2}^{3\times \dfrac{1}{3}}}} \right) \right]}^{2}}$
Now, after solving the power, we will get the value as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}={{\left[ \left( \dfrac{3}{2} \right) \right]}^{2}}$
Here, we will calculate the power of above term as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}=\left( \dfrac{{{3}^{2}}}{{{2}^{2}}} \right)$
Since, the square of $3$ is $9$ and the square of $2$ is 4. So, the above term will become as:
$\Rightarrow {{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}=\left( \dfrac{9}{4} \right)$
Since, the simplified value of the given question is a fraction, we can convert it in decimal also as:
\[\Rightarrow \left( \dfrac{9}{4} \right)=9\div 4=2.25\]
Hence, the simplified value of given question ${{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}$ is \[\left( \dfrac{9}{4} \right)\] or \[2.25\] .

Note: For finding the value of the given question ${{\left( \dfrac{8}{27} \right)}^{\left( -\dfrac{2}{3} \right)}}$, we will have to note that the following points as:
$\Rightarrow $ If the given power is negative, change the power from negative to positive.
$\Rightarrow $ If the power is in fraction as $\dfrac{n}{m}$ , we don’t need to convert the power in decimal. Otherwise we will not be able to simplify the base number.
$\Rightarrow $ The numerator $n$ is number of multiplication of base number to itself and denominator tells to take ${{\text{m}}^{\text{th}}}$ root of the base number.