Question
Download Solution PDFIf \(\rm 7a - {7\over a} + 4 = 0\), then find \(\rm a^3 - {1 \over a^3} - 1\).
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFGiven:
7a - (7/a) = - 4
Formula used:
If a + (1/a) = P
⇒ a3 - (1/a3) = P3 + 3P
Calculation:
7a - (7/a) = - 4
⇒ a - (1/a) = (- 4/7)
⇒ a3 - (1/a3) = (- 4/7)3 + 3 × (- 4/7)
⇒ (- 64/343) - 12/7
⇒ (- 64 - 588)/343
⇒ a3 - (1/a3) = (- 652/343)
a3 - (1/a3) - 1
⇒ - 652/343 - 1
⇒ (- 652 - 343)/343
⇒ - 995/343
∴ The correct answer is (- 995/343).
Last updated on Jun 25, 2025
-> The SSC CGL Notification 2025 has been released on 9th June 2025 on the official website at ssc.gov.in.
-> The SSC CGL exam registration process is now open and will continue till 4th July 2025, so candidates must fill out the SSC CGL Application Form 2025 before the deadline.
-> This year, the Staff Selection Commission (SSC) has announced approximately 14,582 vacancies for various Group B and C posts across government departments.
-> The SSC CGL Tier 1 exam is scheduled to take place from 13th to 30th August 2025.
-> Aspirants should visit ssc.gov.in 2025 regularly for updates and ensure timely submission of the CGL exam form.
-> Candidates can refer to the CGL syllabus for a better understanding of the exam structure and pattern.
-> The CGL Eligibility is a bachelor’s degree in any discipline.
-> Candidates selected through the SSC CGL exam will receive an attractive salary. Learn more about the SSC CGL Salary Structure.
-> Attempt SSC CGL Free English Mock Test and SSC CGL Current Affairs Mock Test.
-> Candidates should also use the SSC CGL previous year papers for a good revision.
->The UGC NET Exam Analysis 2025 for June 25 is out for Shift 1.