Question

prove that (81/16) to the power -3/4× [(25/9) to the power -3/2÷ (5/2) to the power -3] = 1

Answer 1

$$\begin{gathered} \mathrm{L}.\mathrm{H}.\mathrm{S}={\left(\frac{81}{16}\right)}^{-\frac{3}{4}}\times \left[{\left(\frac{25}{9}\right)}^{-\frac{3}{2}}÷{\left(\frac{5}{2}\right)}^{-3}\right] \\ ={\left(\frac{16}{81}\right)}^{\frac{3}{4}}\times \left[{\left(\frac{9}{25}\right)}^{\frac{3}{2}}÷{\left(\frac{2}{5}\right)}^3\right]\mathbf{}\left\{{\left({\displaystyle \frac{\mathit{a}}{\mathit{b}}}\right)}^{\mathbf{-}\mathit{n}}\mathbf{}\mathbf{=}\mathbf{}{\left(\frac{\mathit{b}}{\mathit{a}}\right)}^{\mathit{n}}\right\} \\ ={\left\{{\left(\frac{2}{3}\right)}^4\right\}}^{\frac{3}{4}}\times \left[{\left\{{\left(\frac{3}{5}\right)}^2\right\}}^{\frac{3}{2}}÷{\left(\frac{2}{5}\right)}^3\right]\mathbf{}\mathbf{}\left[{\left({\mathit{a}}^{\mathit{m}}\right)}^{\mathit{n}}\mathbf{}\mathbf{=}\mathbf{}{\mathit{a}}^{\mathit{m}\mathit{n}}\right] \\ ={\left(\frac{2}{3}\right)}^3\times \left[{\left(\frac{3}{5}\right)}^3÷{\left(\frac{2}{5}\right)}^3\right] \\ =\frac{2^3}{3^3}\times \left[\frac{3^3}{5^3}÷\frac{2^3}{5^3}\right] \end{gathered}$$

$$\begin{gathered} =\frac{2^3}{3^3}\times \left[\frac{3^3}{5^3}\times \frac{5^3}{2^3}\right] \\ =\frac{2^3}{3^3}\times \frac{3^3}{2^3}=1=\mathrm{R}.\mathrm{H}.\mathrm{S} \\ \mathbf{}\mathbf{}\mathbf{HENCE}\mathbf{}\mathbf{PROVED} \end{gathered}$$