Question

Evaluate:

${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\times [{\left(\frac{25}{9}\right)}^{-\frac{3}{2}}÷{\left(\frac{5}{9}\right)}^{-3}]$

• A. ${\left(\frac{4}{9}\right)}^3$
• B. ${\left(\frac{2}{3}\right)}^3$
• C. ${\left(\frac{3}{5}\right)}^3$
• D. ${\left(\frac{2}{9}\right)}^3$

Answer 1

The correct option is D ${\left(\frac{2}{9}\right)}^3$

${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\times [{\left(\frac{25}{9}\right)}^{-\frac{3}{2}}÷{\left(\frac{5}{9}\right)}^{-3}]$

$={\left(\frac{3^4}{2^4}\right)}^{-\frac{3}{4}}\times [{\left(\frac{5^2}{3^2}\right)}^{-\frac{3}{2}}\times {\left(\frac{5}{9}\right)}^3]$

$={\left(\frac{3}{2}\right)}^{-3}\times [{\left(\frac{5}{3}\right)}^{-3}\times {\left(\frac{5}{9}\right)}^3]$

$={\left(\frac{2}{3}\right)}^3\times {\left(\frac{3}{5}\right)}^3\times {\left(\frac{5}{9}\right)}^3$

$=\frac{2^3\times 3^{3-3}\times 5^{3-3}}{9^3}$

$={\left(\frac{2\times 1\times 1}{9}\right)}^3$

$={\left(\frac{2}{9}\right)}^3$