Question

If \[x + \frac{1}{x} = 3\]  then find the value of \[x^6 + \frac{1}{x^6}\].

Answer 1

We have to find the value of  `x^6 + 1/x^6`

Given  `x+ 1/x = 3`

Using identity  `(a+b)^2 = a^2 + 2ab +b^2`

Here   `a= x, b=1/x`

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

`(x+1/x)^2 = x^2 + 2 xx x xx 1/x +1/x xx1/x`

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

By substituting the value of   `x+1/x = 3` We get,

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

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

By transposing + 2 to left hand side, we get

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

       `7 = x^2 + 1/x^2`

Cubing on both sides we get,

`(7)^3 = (x^2 + 1/x^2)^3`

Using identity \[\left( a + b \right)^3 = a^3 + b^3 + 3ab\left( a + b \right)\]

Here  `a=x^3 , b=1/x^2`

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

`343 = x^6 + 1/x^6 + 3xx  x^2 xx  1/x^2 (x^2  + 1/x^2)`

Put   `x^2 + 1/x^2 = 7`we get 

`343 = x^6 +1/x^6 + 3 xx 7`

`343 = x^6 +1/x^6 +21`

By transposing 21 to left hand side we get ,

`343 - 21 = x^6 + 1/x^6`

         `322 = x^6 + 1/x^6`

Hence the value of  `x^6 + 1/x^6` is  322.