Question

(8/27)^-2/3 can be simplified to 9/4.... true or false

Answer 1

Remember$x^{-m}=\frac{1}{x^m}$
and
$(\frac{a}{b})^c=\frac{a^c}{b^c}$
and
$x^\frac{m}{n}=\sqrt[n]{x^m}$
and
$(x^m)^n=x^{mn}$

so, combining all of those
$(\frac{8}{27})^\frac{-2}{3}=\frac{8^\frac{-2}{3}}{27^\frac{-2}{3}}$=
$\frac{\frac{1}{8^\frac{2}{3}}}{\frac{1}{27^\frac{2}{3}}}=\frac{27^\frac{2}{3}}{8\frac{2}{3}}$=
$\frac{\sqrt[3]{27^2}}{\sqrt[3]{8^2}}=\frac{\sqrt[3]{(3^3)^2}}{\sqrt[3]{(2^3)^2}}$=
$\frac{\sqrt[3]{3^6}}{\sqrt[3]{2^6}}=\frac{3^ \frac{6}{3}}{2^ \frac{6}{3}}$=
$\frac{3^2}{2^2}=\frac{9}{4}$

true

Answer 2

$\left( \cfrac{8}{27}\right)^{ -\frac{2}{3} } =\left( \cfrac{27}{8}\right)^{ \frac{2}{3} } = \sqrt[3]{\left( \cfrac{27}{8}\right)^{2 } } =\sqrt[3]{ \cfrac{729}{64} } =\sqrt[3]{ \cfrac{9^3}{4^3} } = \cfrac{9}{4}$

True.