Question

Simplify $\frac{{27}^{\frac{2}{3}}}{9^{\frac{1}{2}}{3}^{\frac{3}{2}}}$

• A. $\sqrt{3}$
• B. $1÷\sqrt{3}$
• C. $3\sqrt{3}$
• D. $1÷3\sqrt{3}$

Answer 1

The correct option is B

$1÷\sqrt{3}$

For solving the given expression, $\frac{{27}^{\frac{2}{3}}}{9^{\frac{1}{2}}{3}^{\frac{3}{2}}}$

$27$ is a perfect of $3$, $27$ can be represented as $3\times 3\times 3$ , So the numerator part ${\left(27\right)}^{\frac{2}{3}}$ can be re written as ${\left(3^3\right)}^{\frac{2}{3}}$ which can be further simplified as $3^2$.

In the denominator, we have [${\left(9\right)}^{\frac{1}{2}}{\left(3\right)}^{\frac{3}{2}}$]. Now we know that $9$ is a perfect square and can be represented as $9=3\times 3$.

$9^{\frac{1}{2}}{3}^{\frac{3}{2}}={\left(3^2\right)}^{\frac{1}{2}}{3}^{\frac{\left(3\right)}{2}}={\left(3\right)}^{\frac{5}{2}}$

So, basically we are dividing ${\left(3\right)}^2$ by ${\left(3\right)}^{\frac{5}{2}}$

Using the rule $a^m÷a^n=a^{m-n}$, we get

$\frac{3^2}{3^{\frac{5}{2}}}$ = ${\left(3\right)}^{\frac{-1}{2}}$

= $1÷3^{\frac{1}{2}}$

=$\frac{1}{\sqrt{3}}$