Question

Find the value of ${\left(\frac{1}{216}\right)}^{\frac{-2}{3}}÷{\left(\frac{1}{27}\right)}^{\frac{-4}{3}}$

• A. $\frac{5}{9}$
• B. $\frac{4}{9}$
• C. $\frac{7}{9}$
• D. $\frac{1}{3}$

Answer 1

The correct option is B $\frac{4}{9}$

$(\frac{1}{216}{)}^{\frac{-2}{3}}÷(\frac{1}{27}{)}^{\frac{-4}{3}}$

$(216{)}^{\frac{2}{3}}÷(27{)}^{\frac{4}{3}}$

$(6^3{)}^{\frac{2}{3}}÷(3^3{)}^{\frac{4}{3}}$

$6^2÷3^4$

$36÷81=\frac{4}{9}$

Hence the correct option is (B).