Question

If x² + `x^(1/2)`= 7 and  x ≠ 0; find the value of : 
7x³ + 8x - `7/x^3 - 8/x`

Answer 1

Given that `x^2 + 1/x^2 = 7`

We need to find the value of  7x³ + 8x - `7/x^3 - 8/x`

Consider the given equation :
x2 + `1/x^2` - 2 = 7 - 2  [ subtract 2 from both the sides ]

⇒ `( x - 1/x)^2 = 5`

⇒ `( x - 1/x ) = +- sqrt5`                               ....(1)
∴ `7x^3 + 8x - 7/x^3 - 8/x = 7x^3 - 7/x^3 + 8x - 8/x`

= `7( x^3 - 1/x^3 ) + 8( x - 1/x )`                 .....(2)
Now consider the expansion of `( x - 1/x )^3` :

`( x - 1/x )^3 = x^3 - 1/x^3 - 3( x - 1/x )`

⇒ `x^3 - 1/x^3 = ( x - 1/x )^3 + 3( x - 1/x)`

⇒ `x^3 - 1/x^3 = (sqrt5)^3 + 3(sqrt5)`         ....(3)
Now substitute the value of  `x^3 -  1/x^3` in equation (2), we have
`7x^3 + 8x - 7/x^3 - 8/x = 7( x^3 - 1/x^3) + 8( x - 1/x )`

⇒ `7x^3 + 8x - 7/x^3 - 8/x = 7[(sqrt5)^3 + 3(sqrt5)] + 8[sqrt5]` [From(3)]

⇒ `7x^3 + 8x - 7/x^3 - 8/x = 7[5(sqrt5) + 3(sqrt5)] + 8[sqrt5]]`

⇒ `7x^3 + 8x - 7/x^3 - 8/x = 64[sqrt5]`