Question

${81}^{\frac{1}{4}}\times 9^{\frac{3}{2}}\times {27}^{\frac{-4}{3}}$ is equal to ______.

Answer 1

${81}^{\frac{1}{4}}\times 9^{\frac{3}{2}}\times {27}^{\frac{-4}{3}}$

$=(3^4{)}^{\frac{1}{4}}\times (3^2{)}^{\frac{3}{2}}\times (3^3{)}^{\frac{-4}{3}}$

$=3\times 3^3\times 3^{-4}$

$=3^{(4-4)}$
$=3^0$
$=1$

Hence, the required result is $1$.