Evaluate- - 216 2197 -

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Byju's Answer

Standard VIII

Mathematics

Cube Root Using Prime Factorisation

Evaluate: √...

Question

Evaluate: $\sqrt[3]{\frac{216}{2197}}$ .

\frac{6}{13}

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\frac{7}{13}

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\frac{8}{13}

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\frac{4}{13}

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Solution

The correct option is A $\frac{6}{13}$
On prime factorisation of the numbers individually, we get,
$216 = 2 \times 2 \times 2 - -------- - \times 3 \times 3 \times 3 - -------- - = 2^{3} \times 3^{3} = 6^{3}$ .
$2197 = 13 \times 13 \times 13 - ------------ - = 13^{3}$ .

Then,
$\sqrt[3]{\frac{216}{2197}}$ $= \sqrt[3]{\frac{6^{3}}{13^{3}}} = \frac{6}{13}$ .

Hence, option $A$ is correct.

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Similar questions

Evaluate each of the following: $(i) \frac{2}{3} - \frac{3}{5} (i i) \frac{- 4}{7} - \frac{2}{- 3} (i i i) \frac{4}{7} - \frac{- 5}{- 7} (i v) - 2 - \frac{5}{9} (v) \frac{- 3}{- 8} - \frac{- 2}{7} (v i) \frac{- 4}{13} - \frac{- 5}{26} (v i i) \frac{- 5}{14} - \frac{- 2}{7} (v i i i) \frac{13}{15} - \frac{12}{25} (i x) \frac{- 6}{13} - \frac{- 7}{13} (x) \frac{7}{24} - \frac{19}{36} (x i) \frac{5}{63} - \frac{- 8}{21}$

Q. Using the appropriate properties of operations of rational numbers, evaluate the following:

\frac{3}{13} \times \frac{4}{5} \times \frac{7}{5} \times \frac{4}{13} \times \frac{5}{13}

Q. Evaluate :

27 - {13 + 4 - (8 + 4 - ¯ ¯¯¯¯¯¯¯¯¯¯ ¯ 1 + 3)}

Verify the following :

(i) $\frac{3}{7} \times (\frac{5}{6} + \frac{12}{13}) = (\frac{3}{7} \times \frac{5}{6}) + (\frac{3}{7} \times \frac{12}{13})$
(ii) $\frac{- 15}{4} \times (\frac{3}{7} + \frac{- 12}{5}) = (\frac{- 15}{4} \times \frac{3}{7}) + (\frac{- 15}{4} \times \frac{- 12}{5})$
(iii) $(\frac{- 8}{3} + \frac{- 13}{12}) \times \frac{5}{6} = (\frac{- 8}{3} \times \frac{5}{6}) + (\frac{- 13}{12} \times \frac{5}{6})$
(iv) $\frac{- 16}{7} \times (\frac{- 8}{9} + \frac{- 7}{6}) = (\frac{- 16}{7} \times \frac{- 8}{9}) + (\frac{- 16}{7} \times \frac{- 7}{6})$

Prove that:
(i) $\frac{1}{3 + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{3}} + \frac{1}{\sqrt{3} + 1} = 1$
(ii) $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} + \frac{1}{\sqrt{8} + \sqrt{9}} = 2$

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Evaluate: 3√2162197.

The correct option is A 613On prime factorisation of the numbers individually, we get,216=2×2×2––––––––––×3×3×3––––––––––=23×33=63.2197=13×13×13––––––––––––––=133.Then,3√2162197 =3√63133=613.Hence, option A is correct.

Evaluate: $\sqrt[3]{\frac{216}{2197}}$ .