Question

Evaluate ${\left(\frac{16}{81}\right)}^{-\frac{3}{4}}\times {\left(\frac{49}{9}\right)}^{\frac{3}{2}}+{\left(\frac{343}{216}\right)}^{\frac{2}{3}}$

• A. $\frac{72}{3185}$
• B. $\frac{3185}{72}$
• C. $72$
• D. $300$

Answer 1

The correct option is B $\frac{3185}{72}$

${\left(\right(\frac{16}{81}\left)\right)}^{\left(\frac{-3}{4}\right)}\times {\left(\right(\frac{49}{9}\left)\right)}^{\left(\frac{3}{2}\right)}+{\left(\right(\frac{343}{216}\left)\right)}^{\left(\frac{2}{3}\right)}={\left({\left(\right(\frac{2}{3}\left)\right)}^4\right)}^{\left(\frac{-3}{4}\right)}\times {\left({\left(\right(\frac{7}{3}\left)\right)}^2\right)}^{\left(\frac{3}{2}\right)}+{\left({\left(\right(\frac{7}{6}\left)\right)}^3\right)}^{\left(\frac{2}{3}\right)}={\left(\right(\frac{2}{3}\left)\right)}^{-3}\times {\left(\right(\frac{7}{3}\left)\right)}^3+{\left(\right(\frac{7}{6}\left)\right)}^2={\left(\right(\frac{3}{2}\left)\right)}^3\times {\left(\right(\frac{7}{3}\left)\right)}^3+{\left(\right(\frac{7}{6}\left)\right)}^2={\left(\right(\frac{7}{2}\left)\right)}^3+{\left(\right(\frac{7}{6}\left)\right)}^2=\left(\frac{343}{8}\right)+\left(\frac{49}{36}\right)=\left(\frac{343\times 9+49\times 2}{72}\right)=\left(\frac{3087+98}{72}\right)=\left(\frac{3185}{72}\right)$