Question

The value of ${\left({27}^{\frac{-2}{3}}\right)}^{\frac{1}{2}}\times {\left({64}^{\frac{1}{3}}\right)}^2\times {\left({81}^{\frac{-3}{2}}\right)}^{\frac{1}{6}}$

• A. $\frac{1}{9}$
• B. $\frac{16}{9}$
• C. $\frac{2}{9}$
• D. $-\frac{16}{9}$

Answer 1

The correct option is B $\frac{16}{9}$

${\left({27}^{\frac{-2}{3}}\right)}^{\frac{1}{2}}\times {\left({64}^{\frac{1}{3}}\right)}^2\times {\left({81}^{\frac{-3}{2}}\right)}^{\frac{1}{6}}$

$={\left(\frac{1}{(27{)}^{\frac{2}{3}}}\right)}^{\frac{1}{2}}\times (4^{3\times \frac{1}{3}}{)}^2\times {\left(\frac{1}{(81{)}^{\frac{3}{2}}}\right)}^{\frac{1}{6}}$

$={\left(\frac{1}{27}\right)}^{\frac{2}{3}\times \frac{1}{2}}\times (4{)}^2\times {\left(\frac{1}{81}\right)}^{\frac{3}{2}\times \frac{1}{6}}$

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

$=\frac{1}{3}\times 16\times \frac{1}{3}$

$=\frac{16}{9}$