Question

Find the value of:
$\left[ {{{\left[ {{{\left( {256} \right)}^{ - \left( {{4^{\dfrac{{ - 3}}{2}}}} \right)}}} \right]}^{ - 3}} \div {{\left[ {{4^3} \div {2^{ - 3}}} \right]}^{\dfrac{{ - 1}}{3}}}} \right] = ?$

Answer 1

Hint: To solve this type of question we have to use properties of exponents like ${\left[ {{a^m}} \right]^{\dfrac{1}{n}}} = {\left[ a \right]^{\dfrac{m}{n}}}$ and many more properties.

Complete step-by-step answer:
We have given,
$\left[ {{{\left[ {{{\left( {256} \right)}^{ - \left( {{4^{\dfrac{{ - 3}}{2}}}} \right)}}} \right]}^{ - 3}} \div {{\left[ {{4^3} \div {2^{ - 3}}} \right]}^{\dfrac{{ - 1}}{3}}}} \right]$
We can write it as
${\left[ {{{\left( {{4^4}} \right)}^{ - \left( {{2^{2 \times \dfrac{{ - 3}}{2}}}} \right)}}} \right]^{ - 3}} \div {\left[ {{4^3} \times {2^3}} \right]^{\dfrac{{ - 1}}{3}}}$ $\left[ {\because {{\left( {{a^m}} \right)}^{\dfrac{p}{q}}} = {a^{m \times \dfrac{p}{q}}}} \right]$ and $\left[ {{a^m} \div {b^{ - n}} = \dfrac{{{a^m}}}{{{b^{ - n}}}} = {a^m} \times {b^n}} \right]$
Now we have
${\left[ {{{\left( {{4^4}} \right)}^{ - \left( {{2^{ - 3}}} \right)}}} \right]^{ - 3}} \div {\left[ {{4^3} \times {2^3}} \right]^{\dfrac{{ - 1}}{3}}}$
On solving we get
$ \Rightarrow {\left[ {{{\left( {{4^4}} \right)}^{\dfrac{{ - 1}}{8}}}} \right]^{ - 3}} \div {\left[ {4 \times 2} \right]^{ - 1}}$ $\left[ {\because {a^m} \times {b^m} = {{\left( {a \times b} \right)}^m}} \right]$
Now using the property ${\left[ {{{\left( a \right)}^m}} \right]^{\dfrac{p}{q}}} = {a^{m \times \dfrac{p}{q}}}$ we get,
$
   \Rightarrow {\left( {{4^4}} \right)^{\dfrac{3}{8}}} \times 8 \\
   \Rightarrow {\left( 4 \right)^{\dfrac{3}{2}}} \times 8 = {2^3} \times 8 = 64 \\
$
Hence the required answer is 64.

Note: Whenever we get this type of question the key concept of solving is you have to use your brain as which property of the exponent should apply to make it easier to solve because there is nothing in this type of question other than properties.