Question

Simplifty: $\frac{(64{)}^{-\frac{1}{6}}\times (216{)}^{-\frac{1}{3}}\times (81{)}^{\frac{1}{4}}}{(512{)}^{-\frac{1}{3}}\times (16{)}^{\frac{1}{4}}\times (9{)}^{-\frac{1}{2}}}$

Answer 1

Given $\frac{(64{)}^{-\frac{1}{6}}\times (216{)}^{-\frac{1}{3}}\times (81{)}^{\frac{1}{4}}}{(512{)}^{-\frac{1}{3}}\times (16{)}^{\frac{1}{4}}\times (9{)}^{-\frac{1}{2}}}$
$=\frac{(2^6{)}^{-\frac{1}{6}}\times (6^3{)}^{-\frac{1}{3}}\times (3^4{)}^{\frac{1}{4}}}{(8^3{)}^{-\frac{1}{3}}\times (2^4{)}^{\frac{1}{4}}\times (3^2{)}^{-\frac{1}{2}}}$
$=\frac{(2{)}^{-\frac{1}{6}\times 6}\times (6{)}^{-\frac{1}{3}\times 3}\times (3{)}^{\frac{1}{4}\times 4}}{(8{)}^{-\frac{1}{3}\times 3}\times (2{)}^{\frac{1}{4}\times 4}\times (3{)}^{-\frac{1}{2}\times 2}}$ [since, $(a^m{)}^n=a^{m\times n}$]
$=\frac{2^{-1}\times (2\times 3{)}^{-1}\times 3}{8^{-1}\times 2\times 3^{-1}}$
$=\frac{(2{)}^{-1-1}\times (3{)}^{-1+1}}{{\left(2^3\right)}^{-1}\times 2\times 3^{-1}}$ [since, $a^m\times a^n=a^{m+n}$]
$=\frac{2^{-2}}{2^{-3}\times 2\times 3^{-1}}$ [Since, $a^{-m}=\frac{1}{a^m}$ and $a^0=1$]
$=\frac{3}{2^{-3+1+2}}$
$=\frac{3}{\left(2^0\right)}$
$=3$