Question

Simplify:

$\left(3r^2\right)\times (9r^2{)}^{3/2}÷(27r^{-3}{)}^{1/3}$ and find the power of $r$

Answer 1

We have,
$\left(3r^2\right)\times (9r^2{)}^{3/2}÷(27r^{-3}{)}^{1/3}$

$\Rightarrow \left(3r^2\right)\times \left(\right(3r{)}^2{)}^{3/2}÷(3^3{r}^{-3}{)}^{1/3}$

$\Rightarrow \left(3r^2\right)\times (3r{)}^3÷(3r^{-1})$

$\Rightarrow \frac{\left(3r^2\right)\times (3r{)}^3}{\left(3r^{-1}\right)}$

$\Rightarrow \left(3r^2\right)\times \left(3^3{r}^3\right)\times \left(3^{-1}r\right)$

$\Rightarrow 27r^6$

So, the power of $r$ is $6$.

Hence, this is the answer.