Note that the polynomials $P(x)=(x-a)(x-b)(x-\frac{1}{ab})$ and $Q(x) = (x-c)(x-d)(x-\frac{1}{cd})$ are same because, $$P(x)=(x-a)(x-b)(x-\frac{1}{ab}) = x^3-(a+b+\frac{1}{ab})x^2+(\frac{1}{a}+\frac{1}{b}+ab)x-1 = x^3-(c+d+\frac{1}{cd})x^2+(\frac{1}{c}+\frac{1}{d}+cd)x-1 = Q(x)$$So we get the roots of both polynomials are same. In other words $\{a,b,\frac{1}{ab}\} = \{c,d,\frac{1}{cd}\}$. Which implies that at least two of $a,b,c,d$ are equal. $\blacksquare$