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Question
If `"a" - (1)/"a" = 7`, find `"a"^2 + (1)/"a"^2 , "a"^2 - (1)/"a"^2` and `"a"^3 - (1)/"a"^3`
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Solution
`"a" - (1)/"a" = 7` ...(1)
Squaring both sides of (1),
`("a" - 1/"a")^2` = (7)2
⇒ `"a"^2 + (1)/"a"^2 - 2` = 49
⇒ `"a"^2 + (1)/"a"^2`
= 49 + 2
= 51
Now, `("a" + 1/"a")^2`
= `"a"^2 + (1)/"a"^2 + 2`
= 51 + 2
= 53
⇒ `"a" + (1)/"a"`
= ±`sqrt(53)`
Now `"a"^2 - (1)/"a"^2`
= `("a" + 1/"a")("a" - 1/"a")`
= `(±sqrt(53)) (7)`
= ±7`sqrt(53)`
Cubing both sides of (1),
`("a" - 1/"a")^3` = (7)3
⇒ `"a"^3 - (1)/"a"^3 - 3("a" - 1/"a")` = 343
⇒ `"a"^3 - (1)/"a"^3 - 3(7)` = 343
⇒ `"a"^3 - (1)/("a"^3`
= 343 + 21
= 364.
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