Question

If \[x + \frac{1}{x} = 3\], calculate  \[x^2 + \frac{1}{x^2}, x^3 + \frac{1}{x^3}\] and \[x^4 + \frac{1}{x^4}\]

Answer 1

In the given problem, we have to find the value of  `x^2 + 1/x^2 , x^3 + 1/x^3 , x^4 +1/x^4`

Given  `x+1/x = 3`

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

Here putting `x+1/x = 3`, 

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

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

                 ` 9 = x^2 + 1/x^2 + 2` 

          `9-2 = x^2 + 1/x^2`

                 ` 7 = x^2 + 1/x^2`

Again squaring on both sides we get,

 `(x^2 + 1/x^2)^2 = (7)^2`

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

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

`(7)^2 =x^4 + 1/x^4 + 2 xx x^2 xx 1/x^2`

`49 = x^4 + 1/x^4 + 2`

`49 - 2 = x^4 + 1/x^4`

`47 = x^4 + 1/x^4`

Again cubing on both sides we get,

 `(x+ 1/x)^3 = (3)^3`

We shall use identity  `(a+b)^3 = a^3+ b^3 + 3ab(a+b)`

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

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

                 `27 = x^3 + 1/x^3 + 9`

          `27-9 = x^3 + 1/x^3`

                  ` 18 = x^3 + 1/x^3`

Hence the value of  `x^2 + 1/x^2 ,x^3+ 1/x^3, x^4 + 1/x^4`is 7,18,47 respectively.