Answer 1

`(x^2 + 1/x^2) - 3(x - 1/x) - 2 = 0`

Let `x - 1/x = y` squaring on both sides

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

`=> x^2 + 1/x^2 = y^2 + 2`

Putting these values in the given equation

`(y^2 + 2) - 3y - 2 = 0`

`=> y^2 - 3y = 0`

=> y(y - 3) = 0

if y = 0  or y - 3 = 0

then y = 0   or y = 3

`=> x - 1/x = 0`    or `x - 1/x = 3`

`=> (x^2 - 1)/x = 0    or (x^2 - 1)/x = 3`

`=> x^2 - 1 = 0   or x^2 - 3x - 1 = 0`

`=> (x + 1)(x - 1) = 0  or x = (-(-3) +- sqrt((-3)^2 - 4(1)(-1)))/(2(1))`

=> x = -1 and x = 1    or `x = (3 +- sqrt13)/2`