The answer is 4.

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We seek positive integers $a,b$ , with $b>2$ , such that $ab^3+2b^2+2=2018$ or $ab^3+2b^2=2016=2^5 3^2 7$ . Thus $b^2$ must divide $2016$ ; the only candidates are $b=3,4,6,12$ . We check that all of those "work"; thus the answer is $\boxed{4}$ .