Well, the quantity is ( (x^3 - 2)/x )^7, or (x^3 - 2)^7/ / x^7. So what you want is the coefficient of x^12 in the expansion of the numerator. If you've got the binomial expansion available (and you should have, otherwise it's not a nice problem) the expansion is x^21 - 7(x^18)(2) + (7x6/2)(x^15)(2^2) - ((7x6x5)/(3x2))(x^12)(2^3) + ... so the coefficient is -(35)(8) or -280.
