Question Detail
Question
In the expansion of \((x^2 - \frac{1}{x})^{12}\), the term without x is equal to
Correct Answer
Option C is the correct answer.
Detailed Explanation
General term: T_{r+1} = C(12,r) (x^2)^{12-r} (-1/x)^r = C(12,r) (-1)^r x^{24-2r} x^{-r} = C(12,r) (-1)^r x^{24-3r}. Set exponent = 0: 24 - 3r = 0 ⇒ r = 8. Term = C(12,8) (-1)^8 x^0 = 495. That is not among the options. The options have fractions like 243/28. Possibly the expansion is different. If it were (x^3 - 1/x)^12, then 36-4r=0 ⇒ r=9, coefficient = C(12,9)(-1)^9 = -220. Not matching. Given the options, 243/28 appears, which is 3^5 / 28. This suggests a different binomial.
Hint
Set the exponent of x to zero to find the constant term.