Find the coefficient of \(x^{3}\) in the binomial expansion of \((x - \frac{3}{x^{2}})^{9}\).
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Correct Answer: Option A
Explanation:
\(x - \frac{3}{x^{2}} = x - 3x^{-2}\)
Let the power on x be t, so that the power on \(x^{-2}\) = 9 - t
\((x)^{t}(x^{-2})^{9 - t} = x^{3} \implies t - 18 + 2t = 3\)
\(3t = 3 + 18 = 21 \therefore t = 7\)
To obtain the coefficient of \(x^{3}\), we have
\(^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})\)
= \(\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}\)
\(x - \frac{3}{x^{2}} = x - 3x^{-2}\)
Let the power on x be t, so that the power on \(x^{-2}\) = 9 - t
\((x)^{t}(x^{-2})^{9 - t} = x^{3} \implies t - 18 + 2t = 3\)
\(3t = 3 + 18 = 21 \therefore t = 7\)
To obtain the coefficient of \(x^{3}\), we have
\(^{9}C_{7}(x)^{7}(3x^{-2))^{2} = \frac{9!}{(9 - 7)! 7!}(x)^{7}(9x^{-4})\)
= \(\frac{9 \times 8 \times 7!}{7! 2!} \times 9(x^{3}) = 324x^{3}\)