The correct option is B 6.01 (217)^ 1/3 = (6^3 + 1)^ 1/3 = 6(1 + 1/6^3)^ 1/3 On expansion by binomial theorem = 6(1 + 1/3 × 216 1 × 2/3 × 3 × 2( 1/216)^2 + ...

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Byju's Answer

Standard XI

Mathematics

Numerically Greatest Term

Cube root of ...

Question

Cube root of 217 is

6.01

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6.04

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6.02

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None of these

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Solution

The correct option is A
6.01

$(217)^{\frac{1}{3}}$ = $(6^{3} + 1)^{\frac{1}{3}}$ = $6 (1 + \frac{1}{6^{3}})^{\frac{1}{3}}$

On expansion by binomial theorem

= $6 (1 + \frac{1}{3 \times 216} - \frac{1 \times 2}{3 \times 3 \times 2} (\frac{1}{216})^{2} + . . . . . . . . .)$ = 6.01

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Cube root of 217 is

The correct option is A 6.01 (217)13 = (63+1)13 = 6(1+163)13 On expansion by binomial theorem = 6(1+13×216−1×23×3×2(1216)2+.........) = 6.01

The correct option is A
6.01

$(217)^{\frac{1}{3}}$ = $(6^{3} + 1)^{\frac{1}{3}}$ = $6 (1 + \frac{1}{6^{3}})^{\frac{1}{3}}$

On expansion by binomial theorem

= $6 (1 + \frac{1}{3 \times 216} - \frac{1 \times 2}{3 \times 3 \times 2} (\frac{1}{216})^{2} + . . . . . . . . .)$ = 6.01