Question : What is the value of $\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$?
Option 1: $\mathrm{k}^{16}+\frac{1}{\mathrm{k}^{16}}$
Option 2: $\mathrm{k}^{16}-\frac{1}{\mathrm{k}^{16}}$
Option 3: $\mathrm{k}^8-\frac{1}{\mathrm{k}^8}$
Option 4: $\mathrm{k}^8+\frac{1}{\mathrm{k}^8}$
New: SSC CHSL tier 1 answer key 2024 out | SSC CHSL 2024 Notification PDF
Recommended: How to crack SSC CHSL | SSC CHSL exam guide
Don't Miss: Month-wise Current Affairs | Upcoming government exams
Correct Answer: $\mathrm{k}^8-\frac{1}{\mathrm{k}^8}$
Solution :
$(a+b)(a-b) =a^2-b^2$
$\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}-\frac{1}{\mathrm{k}}\right)\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$
$= (k^2-\frac{1}{k^2})\left(\mathrm{k}^2+\frac{1}{\mathrm{k}^2}\right)\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$
$=(k^4-\frac{1}{k^4})\left(\mathrm{k}^4+\frac{1}{\mathrm{k}^4}\right)$
$= (k^8-\frac{1}{k^8})$
Hence, the correct answer is $(k^8-\frac{1}{k^8})$.
Related Questions
Know More about
Staff Selection Commission Combined High ...
Result | Eligibility | Application | Admit Card | Answer Key | Preparation Tips | Cutoff
Get Updates BrochureYour Staff Selection Commission Combined Higher Secondary Level Exam brochure has been successfully mailed to your registered email id “”.