# [9^(1/3)] x [27^(-1/2)] all divided by [3^(-1/6)] x [3^(-2/3)] ?

Feb 10, 2018

The answer is $1$.

#### Explanation:

Change all the numbers to a common base of $3$, then apply the exponent addition rule:

$\setminus q \quad \setminus \quad \frac{{9}^{\frac{1}{3}} \cdot {27}^{- \frac{1}{2}}}{{3}^{- \frac{1}{6}} \cdot {3}^{- \frac{2}{3}}}$

$= \frac{{\left({3}^{2}\right)}^{\frac{1}{3}} \cdot {\left({3}^{3}\right)}^{- \frac{1}{2}}}{{3}^{- \frac{1}{6}} \cdot {3}^{- \frac{2}{3}}}$

$= \frac{{3}^{\frac{2}{3}} \cdot {3}^{- \frac{3}{2}}}{{3}^{- \frac{1}{6}} \cdot {3}^{- \frac{2}{3}}}$

$= \frac{{3}^{\frac{2}{3} - \frac{3}{2}}}{{3}^{- \frac{1}{6} - \frac{2}{3}}}$

$= \frac{{3}^{\frac{4}{6} - \frac{9}{6}}}{{3}^{- \frac{1}{6} - \frac{4}{6}}}$

$= \frac{{3}^{- \frac{5}{6}}}{{3}^{- \frac{5}{6}}}$

$= {3}^{- \frac{5}{6}} \cdot {3}^{\frac{5}{6}}$

$= {3}^{- \frac{5}{6} + \frac{5}{6}}$

$= {3}^{0}$

$= 1$