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Question
If x > 0 and `x^2 + 1/[9x^2] = 25/36, "Find" x^3 + 1/[27x^3]`
Solution
Given that
`x^2 + 1/[9x^2] = 25/36`
⇒ `x^2 + 1/( 3x )^2 = 25/36` ...(1)
Now consider the expansion of `( x + 1/(3x) )^2` :
`( x + 1/(3x))^2 = x^2 + 1/(3x)^2 + 2 xx x xx 1/(3x)`
= `x^2 + 1/(3x)^2 + 2/3`
= `25/36 + 2/3` [From(1)]
= `[ 25 + 24 ]/36`
= `49/36`
⇒ `x + 1/(3x) = +-sqrt(49/36)`
⇒ `x + 1/(3x) = +- 7/6`
We need to find `x^3 + 1/[27x^3]` :
Let us consider the expansion of `( x + 1/(3x))^3` :
`( x + 1/(3x))^3 = x^3 + 1/[27x^3] + 3 xx x xx 1/[3x] xx (x + 1/(3x))`
⇒ `( 7/6 )^3 = x^3 + 1/[27x^3] + 3 xx x xx 1/[3x] xx ( x + 1/[3x])`
⇒ `[343]/[216] = x^3 + 1/[27x^3] + x + 1/(3x)`
⇒ `[343]/[216] = x^3 + 1/[27x^3] + 7/6` [∵ `x + 1/3x = 7/6` ]
⇒ `([343]/[216] - 7/6) = x^3 + 1/[27x^3]`
⇒ `x^3 + 1/[27x^3] = [(343 - 252)/216]`
⇒ `x^3 + 1/[27x^3] = (91/216)`
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