Question

If x > 0 and `x^2 + 1/[9x^2] = 25/36, "Find"  x^3 + 1/[27x^3]` 

Answer 1

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)`