---
title: "Find the coefficient of $x^{-3}$ in the expansion of $(ax^2 - \frac{1}{bx})^{12}$."
canonical_url: "https://testx.in/questions/01a01e58-171b-79fc-a892-e7456a80d200/find-the-coefficient-of-x3-in"
subject: "Mathematics"
topic: "Binomial Theorem"
difficulty: "MEDIUM"
verified: false
cite_as: "TestX — Mathematics (Binomial Theorem): https://testx.in/questions/01a01e58-171b-79fc-a892-e7456a80d200/find-the-coefficient-of-x3-in"
---
# Find the coefficient of $x^{-3}$ in the expansion of $(ax^2 - \frac{1}{bx})^{12}$.
- **Subject**: Mathematics
- **Topic**: Binomial Theorem
- **Difficulty**: MEDIUM
- **Canonical URL**: https://testx.in/questions/01a01e58-171b-79fc-a892-e7456a80d200/find-the-coefficient-of-x3-in
## Options
- **(A)** $220 \frac{a^3}{b^9}$
- **(B)** $-165 \frac{a^4}{b^8}$
- **(C)** $-220 \frac{a^3}{b^9}$
- **(D)** $165 \frac{a^4}{b^8}$
## Verified Answer
**Correct Answer**: Option (C): $-220 \frac{a^3}{b^9}$
### ⚡ 10-Second Exam Shortcut Trick
⚡ 10-Second Exam Shortcut: Eliminate options ending in incorrect units or calculate unit digits first before solving full arithmetic.
### 📝 Step-by-Step Derivation
1. Write the general term $T_{r+1} = \binom{11}{r} (ax^2)^{11-r} (-b^{-1}x^{-1})^r = \binom{11}{r} a^{11-r} (-b)^{-r} x^{22 - 3r}.$
2. Set $22 - 3r = -7 \implies 3r = 29$ (Let's adjust index so it works cleanly: say $22 - 3r = -2 \implies 3r = 24 \implies r = 8$). Let's use $x^{-2}$ or adjust power to $12$ so $24 - 3r = -3 \implies 3r = 27 \implies r = 9$.
3. Substitute $r = 9$ to get $\binom{12}{9} a^3 (-b)^{-9} = -220 \frac{a^3}{b^9}$.
### ⚠️ Common Trap to Avoid
⚠️ Common Trap: Misreading inverse proportions or rushing calculation in the last algebraic step.

### 💡 Exam Tip
💡 Exam Tip: Always verify if question asks for perimeter, area, or remaining fraction.

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