Simplify the following expression: ${{\left[ {{\left( 64 \right)}^{-2}} \right]}^{-3}}\div {{\left[ {{\left\{ \left( -{{8}^{2}} \right) \right\}}^{3}} \right]}^{2}}$

Hint: First, use the property ${{n}^{-a}}=\dfrac{1}{{{n}^{a}}}$ on the first term of the given expression and ${{\left( -n \right)}^{2}}={{n}^{2}}$ on the second term. Then use ${{n}^{-a}}=\dfrac{1}{{{n}^{a}}}$ again in the first term. Next, use ${{\left( {{n}^{a}} \right)}^{b}}={{n}^{ab}}$ on both the terms of the resulting expression. Finally, divide both the terms of the resulting expression to get the final answer.

Complete step-by-step solution -

In this question, we need to simplify the following expression: ${{\left[ {{\left( 64 \right)}^{-2}} \right]}^{-3}}\div {{\left[ {{\left\{ \left( -{{8}^{2}} \right) \right\}}^{3}} \right]}^{2}}$
We are given the following expression: ${{\left[ {{\left( 64 \right)}^{-2}} \right]}^{-3}}\div {{\left[ {{\left\{ \left( -{{8}^{2}} \right) \right\}}^{3}} \right]}^{2}}$
We know the property that a number n raised to the power of negative of another number a will result in the reciprocal of n to the power of a.
i.e. ${{n}^{-a}}=\dfrac{1}{{{n}^{a}}}$
Also, the square of negative of n is equal to the square of n and both squares are positive.
i.e. ${{\left( -n \right)}^{2}}={{n}^{2}}$
Using these properties in the given expression, we will get the following:
${{\left[ \dfrac{1}{{{\left( 64 \right)}^{2}}} \right]}^{-3}}\div {{\left[ {{\left\{ \left( 64 \right) \right\}}^{3}} \right]}^{2}}$
Now, we will use the first property again on the first term of the equation, we will get the following:
${{\left[ {{\left( 64 \right)}^{2}} \right]}^{3}}\div {{\left[ {{\left\{ \left( 64 \right) \right\}}^{3}} \right]}^{2}}$
Now, we know that if a number n is raised to the power of a number a and this whole number is in turn raised to the power of another number b, then the result is n raised to the power of the product of a and b.
i.e. ${{\left( {{n}^{a}} \right)}^{b}}={{n}^{ab}}$
Using this property in the above equation, we will get the following:
$\left[ {{\left( 64 \right)}^{2\times 3}} \right]\div \left[ {{\left( 64 \right)}^{3\times 2}} \right]$
${{\left( 64 \right)}^{6}}\div {{\left( 64 \right)}^{6}}=1$
Hence, ${{\left[ {{\left( 64 \right)}^{-2}} \right]}^{-3}}\div {{\left[ {{\left\{ \left( -{{8}^{2}} \right) \right\}}^{3}} \right]}^{2}}=1$.
This is our final answer.

Note: In this question, it is very important to know the following properties: a number n raised to the power of negative of another number a will result in the reciprocal of n to the power of a. i.e. ${{n}^{-a}}=\dfrac{1}{{{n}^{a}}}$, the square of negative of n is equal to the square of n and both squares are positive. i.e. ${{\left( -n \right)}^{2}}={{n}^{2}}$, and that if a number n is raised to the power of a number a and this whole number is in turn raised to the power of another number b, then the result is n raised to the power of the product of a and b. i.e. ${{\left( {{n}^{a}} \right)}^{b}}={{n}^{ab}}$