যদি (x² - \(\frac{1}{x^2}\)) = 4\(\sqrt6\) এবং x > 1 হয়, তাহলে (x³ - \(\frac{1}{x^3}\)) এর মান কত?

Question

Question

This question was previously asked in

SSC CGL 2023 Tier-I Official Paper (Held On: 21 Jul 2023 Shift 3)

Answer 1

প্রদত্ত:

(x2 - \(\frac{1}{x^2}\)) = 4\(√6\) = (x + \(\frac{1}{x}\))(x - \(\frac{1}{x}\)) ...(1)

উভয় পক্ষকে বর্গ করলে

⇒ (x² - \(\frac{1}{x^2}\))² = 16 x 6

⇒ x4 + \(\frac{1}{x^4}\) - 2 = 96

⇒ x4 + \(\frac{1}{x^4}\) = 98

⇒ x4 + \(\frac{1}{x^4}\) + 2 = 98 + 2

⇒ (x2 + \(\frac{1}{x^2}\))² = 100

⇒ x2 + \(\frac{1}{x^2}\) = 10

⇒ x2 + \(\frac{1}{x^2}\) + 2 = 10 + 2

⇒ (x + \(\frac{1}{x}\))² = 12

⇒ x + \(\frac{1}{x}\) = √12 = 2√3 ...(2)

সুতরাং

⇒ x - \(\frac{1}{x}\) = 2√2 [সমীকরণ (1) কে (2) দিয়ে ভাগ করে]

এবং

(x - \(\frac{1}{x}\))3 = (x3 - \(\frac{1}{x^3}\)) - 3 (x - \(\frac{1}{x}\))

⇒ (2√2)³ = x3 - \(\frac{1}{x^3}\) - 3(2√2)

⇒ 16√2 + 6√2 = x3 - \(\frac{1}{x^3}\)

⇒ x3 - \(\frac{1}{x^3}\) = 22√2