Question

Simplify: ${\left\{{81}^{-3/4}\times \frac{{16}^{1/4}}{6^{-2}}\times {\left(\frac{1}{27}\right)}^{-4/3}\right\}}^{1/3}$

Answer 1

Step 1: Factorise each terms of the given expression.

$\begin{array}{l}{\left\{{81}^{-3/4}\times \frac{{16}^{1/4}}{6^{-2}}\times {\left(\frac{1}{27}\right)}^{-4/3}\right\}}^{1/3}\\ ={\left\{{\left(3\times 3\times 3\times 3\right)}^{-3/4}\times \frac{{\left(2\times 2\times 2\times 2\right)}^{1/4}}{{\left(3\times 2\right)}^{-2}}\times {\left(\frac{1}{3\times 3\times 3}\right)}^{-4/3}\right\}}^{1/3}\\ ={\left\{{\left(3^4\right)}^{-3/4}\times \frac{{\left(2^4\right)}^{1/4}}{3^{-2}\times 2^{-2}}\times {\left(\frac{1}{3^3}\right)}^{-4/3}\right\}}^{1/3}\end{array}$

Step 2: Simplify the above expression using the laws of rational exponents.

$\begin{array}{l}{\left\{{\left(3^4\right)}^{-3/4}\times \frac{{\left(2^4\right)}^{1/4}}{3^{-2}\times 2^{-2}}\times {\left(\frac{1}{3^3}\right)}^{-4/3}\right\}}^{1/3}\\ ={\left\{{\left(3\right)}^{-4\times \frac{3}{4}}\times \frac{{\left(2\right)}^{4\times \frac{1}{4}}}{3^{-2}\times 2^{-2}}\times {\left(\frac{1}{3}\right)}^{-3\times \frac{4}{3}}\right\}}^{1/3}\left[\because {\left(p^n\right)}^m=p^{m\times n}\right]\\ ={\left\{{\left(3\right)}^{-3}\times \frac{{\left(2\right)}^1}{3^{-2}\times 2^{-2}}\times {\left(\frac{1}{3}\right)}^{-4}\right\}}^{1/3}\\ ={\left\{{\left(3\right)}^{-3-\left(-2\right)-\left(-4\right)}\times 2^{1-\left(-2\right)}\right\}}^{1/3}\left[\because \frac{a^{p/q}}{a^{r/s}}=a^{\left(p/q-r/s\right)}\right]\\ ={\left\{{\left(3\right)}^{-3+2+4}\times 2^{1+2}\right\}}^{1/3}\\ ={\left\{{\left(3\right)}^3\times {\left(2\right)}^3\right\}}^{1/3}\\ ={\left\{{\left(6\right)}^3\right\}}^{1/3}\left[\because a^m\times b^m={\left(ab\right)}^m\right]\\ ={\left(6\right)}^{3\times \frac{1}{3}}\\ =6\end{array}$

Hence, the simplified value of the given expression is $6$.