Question

Solve $[{\left(\frac{-2}{3}\right)}^4\times {\left(\frac{-2}{3}\right)}^2÷{\left(\frac{4}{9}\right)}^3]$

Answer 1

$[{\left(\frac{-2}{3}\right)}^4\times {\left(\frac{-2}{3}\right)}^2]÷{\left(\frac{4}{9}\right)}^3$ = ${\left(\frac{-2}{3}\right)}^6÷{\left(\frac{4}{9}\right)}^3$ [$\because x^m\times x^n=x^{m+n}$] = ${\left(\frac{-2}{3}\right)}^6÷{\left(\frac{2\times 2}{3\times 3}\right)}^3$ = ${\left(\frac{-2}{3}\right)}^6÷{\left[{\left(\frac{-2}{3}\right)}^2\right]}^3$ = $\frac{-2}{3}\times \frac{-2}{3}\times \frac{-2}{3}\times \frac{-2}{3}\times \frac{-2}{3}\times \frac{-2}{3}÷{\left(\frac{2}{3}\right)}^6$ = ${\left(\frac{2}{3}\right)}^6÷{\left(\frac{2}{3}\right)}^6$ = ${\left(\frac{2}{3}\right)}^{6-6}={\left(\frac{2}{3}\right)}^0=1$ [$\because x^m÷x^m=x^0=1$]