The value of $$\root 3 \of {{2^4}\sqrt {{2^{ - 5}}\sqrt {{2^6}} } } $$    is = ?

Solution (By Examveda Team)

$$\eqalign{ & \root 3 \of {{2^4}\sqrt {{2^{ - 5}}\sqrt {{2^6}} } } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( {{2^6}} \right)}^{\frac{1}{2}}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{\left( 2 \right)}^{\left( {6 \times \frac{1}{2}} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 5}}{{.2}^3}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{\left( { - 5 + 3} \right)}}} } {\text{ }} \cr & = \root 3 \of {{2^4}\sqrt {{2^{ - 2}}} } {\text{ }} \cr & = \root 3 \of {{2^4}.{{\left( {{2^2}} \right)}^{\frac{1}{2}}}} \cr & {\text{ = }}\root 3 \of {{2^4}{{.2}^{\left( { - 2 \times \frac{1}{2}} \right)}}} \cr & = \root 3 \of {{2^4}{{.2}^{\left( { - 1} \right)}}} \cr & = \root 3 \of {{2^{\left( {4 - 1} \right)}}} \cr & {\text{ = }}\root 3 \of {{2^3}} \cr & {\text{ = }}{\left( {{2^3}} \right)^{\frac{1}{3}}} \cr & {\text{ = }}{{\text{2}}^{\left( {3 \times \frac{1}{3}} \right)}} \cr & {\text{ = 2 }} \cr} $$