# What is the square root of 27 to the power of 3?

Jul 17, 2015

${\sqrt{27}}^{3} = \sqrt{{27}^{3}} = \sqrt{{3}^{9}} = {3}^{\frac{9}{2}} = {3}^{4} {3}^{\frac{1}{2}} = 81 \sqrt{3}$

#### Explanation:

Use the following identities ($a , b , c \ge 0$):

$\sqrt{a} = {a}^{\frac{1}{2}}$

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

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

Since the question is slightly ambiguous, let me first show that both possible meanings work out the same:

${\sqrt{27}}^{3} = \sqrt{27} \sqrt{27} \sqrt{27} = \sqrt{27 \cdot 27 \cdot 27} = \sqrt{{27}^{3}}$

Now $27 = {3}^{3}$, so

$\sqrt{{27}^{3}} = \sqrt{{\left({3}^{3}\right)}^{3}} = \sqrt{{3}^{3 \cdot 3}} = \sqrt{{3}^{9}}$

Then:

$\sqrt{{3}^{9}} = {\left({3}^{9}\right)}^{\frac{1}{2}} = {3}^{9 \cdot \frac{1}{2}} = {3}^{\frac{9}{2}} = {3}^{4 + \frac{1}{2}} = {3}^{4} {3}^{\frac{1}{2}} = 81 \sqrt{3}$

So: ${\sqrt{27}}^{3} = \sqrt{{27}^{3}} = 81 \sqrt{3}$