Question

If \[x^4 + \frac{1}{x^4} = 119\] , find the value of \[x^3 - \frac{1}{x^3}\]

Answer 1

In the given problem, we have to find the value of  `x^2 - 1/x^3`

Given  `x^4 + 1/x^4 = 119`

We shall use the identity `(x+y)^2 = x^2 + y^2 + 2xy`

Here putting, `x^4 + 1/x^4 = 119`

`(x^2 + 1/x^2)^2 = x^4 + 1/x^4 + 2 xx x^2 xx 1/x^2`

`(x^2 + 1/x^2)^2 = x^4 + 1/x^4 + 2 xx x^2 xx 1/x^2`

`(x^2 + 1/x^3  = x^4 + 1/x^2 + 2`

`(x^2 + 1/x^2)^2=  119 + 2`

`(x^2 + 1/x^2)^2 = 121`

`x^2 + 1/x^2^2  = sqrt(11 xx 11)`

`x^2 + 1/x^2^2  = ±11`

In order to find `(x-1/x)`we are using identity `(x-y)^2 = x^2 + y^2 - 2xy`.

\[\left( x - \frac{1}{x} \right)^2 = x^2 + \frac{1}{x^2} - 2 \times x \times \frac{1}{x}\]​

`(x-1/x)^2 = x^2 + 1/x^2 - 2`

`(x-1/x)^2 =11 - 2`

`(x-1/x)^2 = 9`

`(x-1/x) =sqrt9`

`(x-1/x)=sqrt9`

`(x-1/x) =sqrt(3 xx 3)`

`(x-1/x)= ± 3 `

In order to find `x^3 - 1/x^3` we are using identity  `a^3 - b^3 = (a-b)(a^2 + b^2 + ab)`

`x^3 - 1/x^3 = (x- 1/x)(x^2 + 1/x^2 + x xx 1/x)`
+ x xx )`Here `x^2 + 1/x^2 = 11` and  `(x - 1/x) = 3`

`x^3 - 1/x^3 = (x-1/x)(x^2+ 1/x^2 + x xx 1/x)`

` = 3(11+1)`

` = 3 xx 12`

` = 36`

Hence the value of  `x^3 - 1/x^3`is 36.