The expression $x^3 - 3x^2 - 9x + c$ can be written in the form $(x - a)^2 (x - b)$ if the values of $c$ is

Let $m, n \in N$ and $\operatorname{gcd}(2, n)=1$. If $30\left(\begin{array}{l}30 \\ 0\end{array}\right)+29\left(\begin{array}{l}30 \\ 1\end{array}\right)+\ldots+2\left(\begin{array}{l}30 \\ 28\end{array}\right)+1\left(\begin{array}{l}30 \\ 29\end{array}\right)= n .2^{ m }$ then $n + m$ is equal to (Here $\left.\left(\begin{array}{l} n \\ k \end{array}\right)={ }^{ n } C _{ k }\right)$

Let $\alpha=\sum_{\mathrm{r}=0}^{\mathrm{n}}\left(4 \mathrm{r}^2+2 \mathrm{r}+1\right)^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}$ and $\beta=\left(\sum_{\mathrm{r}=0}^{\mathrm{n}} \frac{{ }^{\mathrm{n}} \mathrm{C}_{\mathrm{r}}}{\mathrm{r}+1}\right)+\frac{1}{\mathrm{n}+1}$. If $140<\frac{2 \alpha}{\beta}<281$ then the value of $n$ is...............

$\left( {\left( {\begin{array}{*{20}{c}}
{21}\\
1
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
1
\end{array}} \right)} \right) + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
2
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
2
\end{array}} \right)} \right)$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
3
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
3
\end{array}} \right)} \right) + \;.\;.\;.$$ + \left( {\left( {\begin{array}{*{20}{c}}
{21}\\
{10}
\end{array}} \right) - \left( {\begin{array}{*{20}{c}}
{10}\\
{10}
\end{array}} \right)} \right) = $

Let $(1 + x)(1 + x + x^2)(1 + x + x^2 + x^3)\,\, ......\,\,$$(1 + x + x^2 + ..... + x^{30}) = $$a_0 + a_1x + a_2x^2$ .....$+$ $a_{465}x^{465}$, then sum of $a_0 + a_2 + a_4 + ......... +$ is

The sum of all the coefficients in the binomial expansion of ${({x^2} + x - 3)^{319}}$ is