Aspire's Library

A Place for Latest Exam wise Questions, Videos, Previous Year Papers,
Study Stuff for MCA Examinations - NIMCET

Previous Year Question (PYQs)



If $x=1+\sqrt[{6}]{2}+\sqrt[{6}]{4}+\sqrt[{6}]{8}+\sqrt[{6}]{16}+\sqrt[{6}]{32}$ then ${\Bigg{(}1+\frac{1}{x}\Bigg{)}}^{24}$ =





Solution

Given:

\[ x = 1 + 2^{1/6} + 4^{1/6} + 8^{1/6} + 16^{1/6} + 32^{1/6} \]

Step 1: Write in powers of \( a = 2^{1/6} \)

\[ x = 1 + a + a^2 + a^3 + a^4 + a^5 = 1 + \frac{a(a^5 - 1)}{a - 1} \]

Step 2: Use identity \( a^6 = 2 \Rightarrow a^5 = \frac{2}{a} \)

\[ x = 1 + \frac{2 - a}{a - 1} = \frac{1}{a - 1} \Rightarrow 1 + \frac{1}{x} = a \Rightarrow \left(1 + \frac{1}{x} \right)^{24} = a^{24} \]

Step 3: Final calculation

\[ a = 2^{1/6} \Rightarrow a^{24} = (2^{1/6})^{24} = 2^4 = \boxed{16} \]

✅ Final Answer: $\boxed{16}$



Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More


Online Test Series,
Information About Examination,
Syllabus, Notification
and More.

Click Here to
View More

Ask Your Question or Put Your Review.

loading...