Advertisements
Advertisements
प्रश्न
If x > 0 and `x^2 + 1/[9x^2] = 25/36, "Find" x^3 + 1/[27x^3]`
Advertisements
उत्तर
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)`
APPEARS IN
संबंधित प्रश्न
Expand : ( x + 8 ) ( x + 10 )
Expand : ( X - 8 ) ( X + 10 )
Expand: `( 2x - 1/x )( 3x + 2/x )`
If a + b + c = p and ab + bc + ca = q ; find a2 + b2 + c2.
If a + `1/a` = m and a ≠ 0; find in terms of 'm'; the value of `a^2 - 1/a^2`.
In the expansion of (2x2 - 8) (x - 4)2; find the value of coefficient of x3.
If a2 + b2 = 34 and ab = 12; find : 3(a + b)2 + 5(a - b)2
If `(x^2 + 1)/x = 3 1/3` and x > 1; Find `x - 1/x`.
Find the value of 'a': 9x2 + (7a - 5)x + 25 = (3x + 5)2
If 3a + 5b + 4c = 0, show that : 27a3 + 125b3 + 64c3 = 180 abc
