If $\left(\frac{3^{6}}{4^{4}}\right) \mathrm{k}$ is the term, independent of $\mathrm{x}$, in the
binomial expansion of $\left(\frac{\mathrm{x}}{4}-\frac{12}{\mathrm{x}^{2}}\right)^{12}$, then $\mathrm{k}$ is equal
to______.
$\left(\frac{x}{4}-\frac{12}{x^{2}}\right)^{12}$
$\mathrm{T}_{\mathrm{r}+1}=(-1)^{\mathrm{r}} \cdot{ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{\mathrm{x}}{4}\right)^{12-\mathrm{r}}\left(\frac{12}{\mathrm{x}^{2}}\right)^{\mathrm{r}}$
$\mathrm{T}_{\mathrm{r}+1}=(-1)^{\mathrm{r}} \cdot{ }^{12} \mathrm{C}_{\mathrm{r}}\left(\frac{1}{4}\right)^{12-\mathrm{r}}(12)^{\mathrm{r}} \cdot(\mathrm{x})^{12-3 \mathrm{r}}$
Term independent of $x \Rightarrow 12-3 r=0 \Rightarrow r=4$
$\mathrm{T}_{5}=(-1)^{4} \cdot{ }^{12} \mathrm{C}_{4}\left(\frac{1}{4}\right)^{8}(12)^{4}=\frac{3^{6}}{4^{4}} \cdot \mathrm{k}$
$\Rightarrow \mathrm{k}=55$