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

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Answer 1 · 2015-07-17T18:29:24

$sqrt(27)^3 = sqrt(27^3) = sqrt(3^9) = 3^(9/2) = 3^4 3^(1/2) = 81sqrt(3)$

Use the following identities ( $a, b, c>= 0$ ):

$sqrt(a) = a^(1/2)$

$(a^b)^c = a^(bc)$

$a^(b+c) = a^b a^c$

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

$sqrt(27)^3 = sqrt(27)sqrt(27)sqrt(27) = sqrt(27*27*27) = sqrt(27^3)$

Now $27 = 3^3$ , so

$sqrt(27^3) = sqrt((3^3)^3) = sqrt(3^(3*3)) = sqrt(3^9)$

Then:

$sqrt(3^9) = (3^9)^(1/2) = 3^(9*1/2) = 3^(9/2) = 3^(4+1/2) = 3^4 3^(1/2) = 81sqrt(3)$

So: $sqrt(27)^3 = sqrt(27^3) = 81sqrt(3)$