Question
Download Solution PDFp – 1/p = √7, అయితే p3 – 1/p3 విలువ ఎంత?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFఇచ్చినది:
p – 1/p = √7
ఫార్ములా:
P3 – 1/p3 = (p – 1/p)3 + 3(p – 1/p)
లెక్కింపు:
P3 – 1/p3 = (p – 1/p)3 + 3 (p – 1/p)
⇒ p3 – 1/p3 = (√7)3 + 3√7
⇒ p3 – 1/p3 = 7√7 + 3√7
⇒ p3 – 1/p3 = 10√7
చిన్న ఉపాయం
x - 1/x = a, అప్పుడు x3 - 1/x3 = a3 + 3a
ఇక్కడ, a = √5
ఇక్కడ,
p3 – 1/p3 = (√7)3 + 3 × √7 = 7√7 + 3√7 = 10√7.
Last updated on Jul 9, 2025
-> RRB NTPC Under Graduate Exam Date 2025 has been released on the official website of the Railway Recruitment Board.
-> Bihar Police Admit Card 2025 has been released at csbc.bihar.gov.in
-> The RRB NTPC Admit Card will be released on its official website for RRB NTPC Under Graduate Exam 2025.
-> Candidates who will appear for the RRB NTPC Exam can check their RRB NTPC Time Table 2025 from here.
-> The RRB NTPC 2025 Notification released for a total of 11558 vacancies. A total of 3445 Vacancies have been announced for Undergraduate posts like Commercial Cum Ticket Clerk, Accounts Clerk Cum Typist, Junior Clerk cum Typist & Trains Clerk.
-> A total of 8114 vacancies are announced for Graduate-level posts in the Non-Technical Popular Categories (NTPC) such as Junior Clerk cum Typist, Accounts Clerk cum Typist, Station Master, etc.
-> Prepare for the exam using RRB NTPC Previous Year Papers.
-> Get detailed subject-wise UGC NET Exam Analysis 2025 and UGC NET Question Paper 2025 for shift 1 (25 June) here