Question 2: \((ii)\) \(\quad\) \(\large{\left(\frac{3}{4}\right)^{-2} \div \left(\frac{4}{9}\right)^{3} \times \frac{16}{27}} \)
Solution:
\[\begin{align*}
&=\left(\frac{3}{4}\right)^{-2} \div \left(\frac{4}{9}\right)^{3} \times \frac{16}{27}\\
&=\left(\frac{4}{3}\right)^{2} \times \left(\frac{4}{9}\right)^{-3} \times \frac{16}{27}\\
&=\frac{4^2}{3^2} \times \left(\frac{9}{4}\right)^{3} \times \frac{16}{27}\\
&=\frac{16}{9} \times \frac{9^3}{4^3} \times \frac{16}{27}\\
&=\frac{16}{9} \times \frac{729}{64} \times \frac{16}{27}\\
&=12\\
\end{align*}\]
——– Alternate Method ——–
\[\begin{align*}
&=\left(\frac{3}{4}\right)^{-2} \div \left(\frac{4}{9}\right)^{3} \times \frac{16}{27}\\
&=\left(\frac{4}{3}\right)^{2} \times \left(\frac{4}{9}\right)^{-3} \times \frac{16}{27}\\
&=\frac{4^2}{3^2} \times \left(\frac{4}{3^2}\right)^{-3} \times \frac{4^2}{3^3}\\
&=\frac{4^2}{3^2} \times \frac{4^{-3}}{3^{2\times -3}} \times \frac{4^2}{3^3}\\
&=\frac{4^2}{3^2} \times \frac{4^{-3}}{3^{-6}} \times \frac{4^2}{3^3}\\
&=\frac{4^{2-3+2}}{3^{2-6+3}}\\
&=\frac{4}{3^{-1}}\\&=4 \times 3\\
&=12 \\
\end{align*}\]