Question

If \[x^2 + \frac{1}{x^2} = 98\] ,find the value of \[x^3 + \frac{1}{x^3}\]

Answer 1

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

Given   `x^3 + 1/x^3 = 98`

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

Here putting `x^2 + 1/x^2 = 98`,

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

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

`(x+1/x)^2 = 98 + 2`

`(x+1/x)^2 = 100`

`(x+1/x) = sqrt100`

`(x+1/x) = ± 10`

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)`

Here  `(x+1/x) = 10` and `x^2 + 1/x^2 = 98`

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

` = 10 (98 - 1)`

` = 10 xx 97`

` = 970`

Hence the value of  `x^3 + 1/x^3` is 970.