# How do you simplify (27 ^(1/12) * 27 ^(-5/12))^-2?

Jul 21, 2016

$9$

#### Explanation:

Keep in mind that

${\left({a}^{b}\right)}^{c} = {a}^{b \cdot c}$

and

${a}^{b} \cdot {a}^{c} = {a}^{b + c}$

and

${a}^{\frac{b}{c}} = \sqrt[c]{{a}^{b}}$

and

$\sqrt[b]{{a}^{b}} = a$

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Use the first concept above and apply it to the first step.

${\left({27}^{\frac{1}{12}} \cdot {27}^{- \frac{5}{12}}\right)}^{-} 2$

$\left({27}^{\frac{1}{12} \cdot - \frac{2}{1}} \cdot {27}^{- \frac{5}{12} \cdot - \frac{2}{1}}\right)$

$\left({27}^{- \frac{2}{12}} \cdot {27}^{\frac{10}{12}}\right)$

Simplify the fractions that are serving as exponents.

$\left({27}^{- \frac{1}{6}} \cdot {27}^{\frac{5}{6}}\right)$

Now apply the second concept mentioned above. Simplify the fractions again.

$\left({27}^{- \frac{1}{6} + \frac{5}{6}}\right)$

${27}^{\frac{4}{6}}$

${27}^{\frac{2}{3}}$

Use the third concept noted at the top.

$\sqrt[3]{{27}^{2}}$

Calculate the the value inside the radicand and rewrite.

$\sqrt[3]{729}$

Find the cube root by rewriting the radicand again. Follow the fourth concept after.

$\sqrt[3]{{9}^{3}}$

$9$