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Question
If `"a" + (1)/"a" = "p"`; then show that `"a"^3 + (1)/"a"^3 = "p"("p"^2 - 3)`
Sum
Solution
`"a" + (1)/"a" = "p"`
`("a" + 1/"a")^3`
= `"a"^3 + (1)/"a"^3 + 3("a" + 1/"a")`
⇒ p3 = `"a"^3 + (1)/"a"^3 + 3("p")`
⇒ `"a"^3 + (1)/"a"^3`
= p3 - 3p
= p(p2 - 3).
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