How do you simplify the expression: \[\dfrac{{{\left( {{a}^{8}}{{a}^{6}} \right)}^{\dfrac{1}{7}}}}{{{a}^{2}}}\]?

Hint: We first explain the process of exponents and indices. We find the general form for the identities. Then we perform those identities on exponents. We find the solution of the fraction form separately for denominator and denominator.

Complete step by step solution:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$.
In case the value of $n$ becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$.
For our given fraction \[\dfrac{{{\left( {{a}^{8}}{{a}^{6}} \right)}^{\dfrac{1}{7}}}}{{{a}^{2}}}\], we simplify the numerator part first.
If we take two exponential expressions where the exponents are $m$ and $n$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. So, \[{{a}^{8}}{{a}^{6}}={{a}^{8+6}}={{a}^{14}}\].
We also have the identity of ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$. Putting the values, we get
\[{{\left( {{a}^{8}}{{a}^{6}} \right)}^{\dfrac{1}{7}}}={{\left( {{a}^{14}} \right)}^{\dfrac{1}{7}}}={{a}^{14\times \dfrac{1}{7}}}={{a}^{2}}\].
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
Therefore, \[\dfrac{{{\left( {{a}^{8}}{{a}^{6}} \right)}^{\dfrac{1}{7}}}}{{{a}^{2}}}=\dfrac{{{a}^{2}}}{{{a}^{2}}}=1\].
The simplified form of \[\dfrac{{{\left( {{a}^{8}}{{a}^{6}} \right)}^{\dfrac{1}{7}}}}{{{a}^{2}}}\] is 1.

Note:
All the indices can be performed all together. The base for every term is $a$. Therefore, we can use the operations all at one time to find the solution. The multiplication and division will work as addition and subtraction.