Question

The value of (-3/8)^-3×(4/9)^-2 is

Answer 1

$(\frac{ - 3}{8})^{ - 3} \times (\frac{4}{9}) ^{ - 2} \\ =( \frac{ - 8}{3} )^{3} \times ( \frac{9}{4}) ^{2} \\ = \frac{ - 8}{3} \times \frac{ - 8}{3} \times \frac{ - 8}{3} \times \frac{9}{4} \times \frac{9}{4} \\ = - 96$

Answer 2

$Given \: \Big( \frac{-3}{8}\Big)^{-3} \times \Big( \frac{4}{9}\Big)^{-2}$

$= \Big( \frac{8}{-3}\Big)^{3} \times \Big( \frac{9}{4}\Big)^{2}$

$\boxed { \pink { Since, \Big(\frac{a}{b}\Big)^{-n} = \Big( \frac{b}{a}\Big)^{n} }}$

$= \Big( \frac{(2^{3})^{3}}{(-3)^{3}}\Big)\times \Big( \frac{(3^{2})^{2}}{(2^{2})^{2}}\Big)$

$= \frac{2^{9}}{-3^{3}} \times \frac{3^{4}}{2^{4}}$

$= - 2^{9-4} \times 3^{4-3}$

$\boxed { \pink { \frac{a^{m}}{a^{n}} = a^{m-n}}}$

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

$= - 32 \times 3 \\= - 96$

Therefore.,

$\red { \Big( \frac{-3}{8}\Big)^{-3} \times \Big( \frac{4}{9}\Big)^{-2}}\green { = -96}$

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