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By the Binomial Theorem, [2x-(1/2x)]^6 can be expanded to the sum as k goes from zero to 6 of (6 choose k)×(2x)k×(-1/2x)6-k. The only time we will be left with an x2 term is when k=4, since we'll have some coefficient
multiplied by x4/x2. So, we'll have (6 choose 4) × (2x)4×(-1/2x)2,...

(I'm assuming this is meant to be read as "...for which x-1 fails to exist...", and I'm also assuming that this is under the operation of multiplication.)
For any element A of Z15 to have an inverse, that inverse B must be such that
AB = BA = 1...