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If the coefficient of $x^3$ in the binomial expansion of $x^3\left(2 \sqrt{3} x^2+\frac{1}{k x}\right)^{12}$ is 880 , then $k$ is equal to
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$\sqrt{3}$
$\begin{aligned} & x^3\left[{ }^{12} C_4\left(2 \sqrt{3} x^2\right)^4 \cdot\left(\frac{1}{k x}\right)^8\right]=880 x^3 \\ & \Rightarrow \quad \frac{12 !}{4 ! 8 !} \times(16 \times 9) \times \frac{1}{k^8}=880\end{aligned}$
$\begin{array}{rlrl}\Rightarrow & \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times 9 \times \frac{1}{k^8} & =5.5 \\ \Rightarrow & 11 \times 5 \times 9 \times 9 & =55 \times k^8 \\ \Rightarrow & k^8=81 \Rightarrow k=\sqrt{3}\end{array}$
$\begin{array}{rlrl}\Rightarrow & \frac{12 \times 11 \times 10 \times 9}{4 \times 3 \times 2} \times 9 \times \frac{1}{k^8} & =5.5 \\ \Rightarrow & 11 \times 5 \times 9 \times 9 & =55 \times k^8 \\ \Rightarrow & k^8=81 \Rightarrow k=\sqrt{3}\end{array}$
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