Question

8 by 27 to the power minus 1 by 3 into 25 by 4 to the power 1 by 2 into 4 by 9 to the power 0 + 125 by 64 to the power 1 by 3​

Answer 1

$\displaystyle(\frac{8}{27})^\frac{-1}{3}{\times}(\frac{25}{4})^\frac{1}{2}{\times}(\frac{4}{9})^0+(\frac{125}{64})^\frac{1}{3}$

$\text{Using, }$

$\boxed{\bf\;a^0=1}\;\boxed{\bf\;(\frac{a}{b})^{-m}=(\frac{b}{a})^{m}}$

$=\displaystyle(\frac{27}{8})^\frac{1}{3}{\times}(\frac{25}{4})^\frac{1}{2}{\times}1+(\frac{125}{64})^\frac{1}{3}$

$=\displaystyle((\frac{3}{2})^3)^\frac{1}{3}{\times}((\frac{5}{2})^2)^\frac{1}{2}+((\frac{5}{4})^3)^\frac{1}{3}$

$\text{Using,}$

$\boxed{\bf(a^m)^n=a^{mn}}$

$=\displaystyle\frac{3}{2}{\times}\frac{5}{2}+\frac{5}{4}$

$=\displaystyle\frac{15}{4}+\frac{5}{4}$

$=\displaystyle\frac{20}{4}$

$=\displaystyle\;5$