(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