If \(x+\frac{1}{x}\) = p, then \(x^6+\frac{1}{x^6}\) equals to :(a) p6

1.	If \(x+\frac{1}{x}\) = p, then \(x^6+\frac{1}{x^6}\) equals to :(a) p6 + 6p (b) p6 – 6p (c) p6 + 6p4 + 9p2 + 2 (d) p6 – 6p4 + 9p2 – 2
Answer» (d) p⁶ – 6p⁴ + 9p² – 2 Given, \(x+\frac{1}{x}\) = p ⇒ \(\bigg(x+\frac{1}{x}\bigg)^2\) = p² ⇒ \(x^2+\frac{1}{x^2}+2 = p^2\) ⇒ \(x^2+\frac{1}{x^2} = p^2 - 2\) ⇒ \(\bigg(x^2+\frac{1}{x^2}\bigg)^3\) = (p² - 2)³ ⇒ \(x^6+\frac{1}{x^6}\) + 3\(\bigg(x^2+\frac{1}{x^2}\bigg)\) = p⁶ - 8 + 6p² (p² - 2) ⇒ \(x^6+\frac{1}{x^6}\) + 3 (p² - 2) = p⁶ - 8 + 6p² (p² - 2) ⇒ \(x^6+\frac{1}{x^6}\) = p⁶ - 6p⁴- 9p² - 2

If \(x+\frac{1}{x}\) = p, then \(x^6+\frac{1}{x^6}\) equals to :(a) p6 + 6p (b) p6 – 6p (c) p6 + 6p4 + 9p2 + 2 (d) p6 – 6p4 + 9p2 – 2

Answer»

(d) p⁶ – 6p⁴ + 9p² – 2

Given, \(x+\frac{1}{x}\) = p ⇒ \(\bigg(x+\frac{1}{x}\bigg)^2\) = p²

⇒ \(x^2+\frac{1}{x^2}+2 = p^2\) ⇒ \(x^2+\frac{1}{x^2} = p^2 - 2\)

⇒ \(\bigg(x^2+\frac{1}{x^2}\bigg)^3\) = (p² - 2)³

⇒ \(x^6+\frac{1}{x^6}\) + 3\(\bigg(x^2+\frac{1}{x^2}\bigg)\) = p⁶ - 8 + 6p² (p² - 2)

⇒ \(x^6+\frac{1}{x^6}\) + 3 (p² - 2) = p⁶ - 8 + 6p² (p² - 2)

⇒ \(x^6+\frac{1}{x^6}\) = p⁶ - 6p⁴- 9p² - 2