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The sum of the coefficients of the first three terms in the expansion of $(x - \frac{3}{x^2})^m$,where $x \neq 0$ and $m$ is a natural number,is $559$. Find the term of the expansion containing $x^3$. (in $x^3$)

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If the $7^{th}$ term in the binomial expansion of $\left( \frac{3}{\sqrt[3]{84}} + \sqrt{3} \ln x \right)^9, x > 0,$ is equal to $729,$ then $x$ can be

If the constant term in the expansion of $\left(\frac{\sqrt[5]{3}}{x}+\frac{2x}{\sqrt[3]{5}}\right)^{12}, x \neq 0$,is $\alpha \times 2^8 \times \sqrt[5]{3}$,then $25 \alpha$ is equal to :

If $(1-x+x^2)^{10}=a_0+a_1 x+a_2 x^2+\ldots+a_{20} x^{20}$,then $2 a_2+3 a_3+4 a_4+\ldots+20 a_{20}=$

If the $k^{\text{th}}$ term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^6$ is independent of $x$,then the numerically greatest term in the expansion of $\left(\frac{3}{2} x^2 - \frac{1}{3x}\right)^k$ when $x = \frac{2}{3}$ is:

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