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Question
If `"a" + (1)/"a" = 2`, then show that `"a"^2 + (1)/"a"^2 = "a"^3 + (1)/"a"^3 = "a"^4 + (1)/"a"^4`
Solution
`"a" + (1)/"a" = 2`
`("a" + 1/"a")^2`
= `"a"^2 + (1)/"a"^2 + 2`
⇒ (2)2 = `"a"^2 + (1)/"a"^2 + 2`
⇒ `"a"^2 + (1)/"a"^2`
= 4 - 2
= 2
`("a" + 1/"a")^3`
= `"a"^3 + (1)/"a"^3 + 3("a" + 1/"a")`
⇒ (2)3 = `"a"^3 + (1)/"a"^3 + 3(2)`
⇒ `"a"^3 + (1)/"a"^3`
= 8 - 6
= 2
`("a"^2 + 1/"a"^2)^2`
= `"a"^4 + (1)/"a"^4 + 2`
⇒ (2a)2 = `"a"^4 + (1)/"a"^4 + 2`
⇒ `"a"^4 + (1)/"a"^4`
= 4 - 2
= 2
Thus, `"a"^2 + (1)/"a"^2 = "a"^3 + (1)/"a"^3 = "a"^4 + (1)/"a"^4`
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