Coefficient X3 B63D2C

1. **State the problem:** We need to find the values of $a$ such that the coefficient of $x^3$ in the expansion of $$\left(2x + \frac{1}{ax^2}\right)^9$$ is 288. 2. **Recall the binomial expansion formula:** For any $n$, $\left(p + q\right)^n = \sum_{k=0}^n \binom{n}{k} p^{n-k} q^k$. 3. **Identify terms:** Here, $p = 2x$ and $q = \frac{1}{ax^2}$. 4. **General term:** The $k$-th term is $$T_{k+1} = \binom{9}{k} (2x)^{9-k} \left(\frac{1}{ax^2}\right)^k = \binom{9}{k} 2^{9-k} x^{9-k} \frac{1}{a^k x^{2k}} = \binom{9}{k} \frac{2^{9-k}}{a^k} x^{9-k-2k} = \binom{9}{k} \frac{2^{9-k}}{a^k} x^{9-3k}.$$ 5. **Find $k$ for which the power of $x$ is 3:** $$9 - 3k = 3 \implies 3k = 6 \implies k = 2.$$ 6. **Find the coefficient for $k=2$:** $$\binom{9}{2} \frac{2^{9-2}}{a^2} = \binom{9}{2} \frac{2^7}{a^2} = 36 \times \frac{128}{a^2} = \frac{4608}{a^2}.$$ 7. **Set the coefficient equal to 288 and solve for $a$:** $$\frac{4608}{a^2} = 288 \implies a^2 = \frac{4608}{288} = 16.$$ 8. **Solve for $a$:** $$a = \pm 4.$$ **Final answer:** The possible values of $a$ are $\boxed{4}$ and $\boxed{-4}$.