If the coefficient of x^3 in the binomial exp | vedclass

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(D) The general term in the expansion of $(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $T_{r+1} = {}^{12}C_r (2 \sqrt{3} x^2)^{12-r} (\frac{1}{kx})^r$.The

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$\sqrt{3}$

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(D) The general term in the expansion of $(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $T_{r+1} = {}^{12}C_r (2 \sqrt{3} x^2)^{12-r} (\frac{1}{kx})^r$.The expression is $x^3 	imes T_{r+1} = {}^{12}C_r (2 \sqrt{3})^{12-r} (\frac{1}{k})^r x^{3 + 2(12-r) - r} = {}^{12}C_r (2 \sqrt{3})^{12-r} (\frac{1}{k})^r x^{27-3r}$.For the coefficient of $x^3$,we set $27 - 3r = 3$,which gives $3r = 24$,so $r = 8$.The coefficient is ${}^{12}C_8 (2 \sqrt{3})^{12-8} (\frac{1}{k})^8 = {}^{12}C_4 (2 \sqrt{3})^4 (\frac{1}{k^8}) = 880$.${}^{12}C_4 = \frac{12 	imes 11 	imes 10 	imes 9}{4 	imes 3 	imes 2 	imes 1} = 495$.$(2 \sqrt{3})^4 = 16 	imes 9 = 144$.So,$495 	imes 144 	imes \frac{1}{k^8} = 880$.$\frac{71280}{k^8} = 880 \Rightarrow k^8 = \frac{71280}{880} = 81$.$k^8 = 3^4$ $\Rightarrow k^2 = \sqrt{3}$ $\Rightarrow k = \sqrt[4]{3}$ is incorrect based on the provided options; re-evaluating: $k^8 = 81 = 3^4$,so $k = (3^4)^{1/8} = 3^{1/2} = \sqrt{3}$.

If the coefficient of $x^3$ in the binomial expansion of $x^3(2 \sqrt{3} x^2 + \frac{1}{kx})^{12}$ is $880$,then $k$ is equal to

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