Notice that
(1 - x)⁵ (1 + 1/x)⁵ = ((1 - x) (1 + 1/x))⁵ = (1 - x + 1/x - 1)⁵ = (1/x - x)⁵
Recall the binomial theorem:
[tex]\displaystyle(a+b)^n = \sum_{k=0}^n\binom nk a^{n-k}b^k[/tex]
Let a = 1/x, b = -x, and n = 5. Then
[tex]\displaystyle\left(\frac1x-x\right)^5 = \sum_{k=0}^5\binom5k\left(\frac1x\right)^{5-k}(-x)^k = \sum_{k=0}^5 (-1)^k\binom5k x^{2k-5}[/tex]
We get an x ³ term for
2k - 5 = 3 ==> 2k = 8 ==> k = 4
so that the coefficient would be
[tex](-1)^4\dbinom54 = \boxed{5}[/tex]
