x3 + x2 – (1/x2) + (1/x3)
Rearranging the above terms, we get,
x3 + (1/x3) + x2 – (1/x2)
We know that, a3 – b3 = (a – b) (a2 + ab + b2) and (a2 – b2) = (a + b) (a – b)
(x + 1/x) (x2 – 1 + 1/x2) + (x + 1/x) (x – 1/x)
(x + 1/x) [x2 – 1 + 1/x2 + x – 1/x]