The cube root of $27^{2}$ is __________.

27

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9

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3

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None of these

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Solution

The correct option is A $9$
On prime factorising, we get,
$27 = 3 \times 3 \times 3 - -------- - = 3^{3}$ .

Then, cube root of $27^{2}$ is:
$\sqrt[3]{27^{2}} = \sqrt[3]{(3^{3})^{2}} = 3^{2} = 9$ .
Therefore, option $B$ is correct.

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Similar questions

27^{2}

Two cubes have their volumes in the ratio $1 : 27$ . The ratio of their surface areas is

(a) $1 : 3$

(b) $1 : 9$

(d) none of these

^{'} ω^{'}

ω^{(\frac{1}{3} + \frac{2}{9} + \frac{4}{27} + . . . \infty)} + ω^{(\frac{1}{2} + \frac{3}{8} + \frac{9}{32} + . . . \infty)} =

ω

ω^{⎛ ⎝ \frac{1}{3} + \frac{2}{9} + \frac{4}{27} + . . . \infty ⎞ ⎠} + ω^{⎛ ⎝ \frac{1}{2} + \frac{3}{8} + \frac{9}{32} + . . . \infty ⎞ ⎠} =

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