Binomial Expansion

Question

The term independent of x in the expansion of (x² - 1/x)⁹ is

RankGuru · Accepted Answer

1. **General Term**: For $(x^2 - x^{-1})^9$, the general term is:
   - $T_{r+1} = \binom{9}{r} (x^2)^{9-r} (-x^{-1})^r$
   - $T_{r+1} = \binom{9}{r} (x)^{18-2r} (-1)^r (x)^{-r} = \binom{9}{r} (-1)^r x^{18-3r}$
2. **Independent Condition**: For the term to be independent of $x$, the power of $x$ must be zero:
   - $18 - 3r = 0 \implies r = 6$.
3. **Calculate Value**:
   - $T_7 = \binom{9}{6} (-1)^6 x^0 = \binom{9}{3} \cdot 1$
   - $T_7 = \frac{9 	imes 8 	imes 7}{3 	imes 2 	imes 1} = 3 	imes 4 	imes 7 = 84$.

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