Question

If \[x^4 + \frac{1}{x^4} = 194,\] find \[x^3 + \frac{1}{x^3}, x^2 + \frac{1}{x^2}\] and \[x + \frac{1}{x}\]

Answer 1

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

Given  `x^4 + 1/x^4 = 194`

By adding and subtracting  `2 xx x^2 xx 1/x^2`in left hand side of  `x^4 + 1/x^4 = 194` we get,

`x^4 + 1/x^4 + 2 xx x^2 xx 1/x^2 -2 xx x^2 xx 1/x^2 = 194`

`x^4 + 1/x^4 + 2 xx x^2 xx 1/x^2 -2 xx (x^2 xx 1/x^2 )= 194`

                                            `(x^2 xx 1/x^2 )^2  - 2= 194`

                                            `(x^2 xx 1/x^2 )^2  - 2= 194 + 2`

                                            `(x^2 xx 1/x^2 )^2  - 2= (14)^2`

                                            `(x^2 xx 1/x^2 )^2  - 2= 14^2`

Again by adding and subtracting  `2xx x xx1/x`in left hand side of   `(x^2 + 1/x^3) = 14`we get,

`x^2 + 1/x^2 + 2xx x xx 1/x -2 xx x xx 1/x =14`

`(x+ 1/x)^2 -2 xx x xx 1/x = 14`

`(x+ 1/x)^2 -2 = 14`

`(x+ 1/x)^2 = 14+ 2`

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

`(x+1/x) = 4`

Now cubing on both sides of  `(x+1/x) = 4` we get

 `(x+1/x)^3 = 4^3`

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

`x^3 + 1/x^3 + 3 xx x xx 1/x (x+1/x) = 4 xx 4xx 4`

`x^3 + 1/x^2 + 3 xx x xx 1/x xx 4 = 64`

`x^3 + 1/x^2 + 12 = 64`

`x^3 + 1/x^2 + 12 = 64 - 12`

`x^3 + 1/x^3  = 52`

Hence the value of  `x^2 + 1/x^2 ,x^2 + 1/x^2 , x+ 1/x` is 52,14,4 respectively.