Question

$(9{)}^{-3}\times \frac{{16}^{\frac{1}{4}}}{6^{-2}}\times {\left(\frac{1}{27}\right)}^{-\frac{4}{3}}$ = ?

• A. $2$
• B. $4$
• C. $6$
• D. $8$

Answer 1

The correct option is D $8$

$9^{-3}\times \frac{{16}^{\frac{1}{4}}}{6^{-2}}\times {\left(\frac{1}{27}\right)}^{\frac{-4}{3}}$

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

$=3^{-6}\times 2\times 6^2\times 3^4$

$=3^{-6}\times 2\times {(2\times 3)}^2\times 3^4$

$=3^{-6}\times 2\times 2^2\times 3^2\times 3^4$ using ${\left(ab\right)}^m=a^m{b}^m$

$=3^{-6+2+4}\times 2^{1+2}=2^3=8$