The coefficient of ; $x ^ { 8 }$& ; in the expansion of  $\left( 1 + x ^ { 4 } \right) ^ { 3 } ( 1 - x ) ^ { 12 }$  is

$\rightarrow$ If $a$ and $b$ are real numbers and $n$ is a positive integer,

then $(a+b)^{n}=n C_{0} a^{n}+n C_{1} a^{n-1} b^{1}+n C_{2} a^{n-2} b^{2}+n C_{3} a^{n-3} b^{3}$

$+\ldots \ldots+n C_{r} a^{n-r} b^{r}+\ldots+n C_{n} b^{n}$

where ${ }^{n} C_{r}=\frac{n !}{r !(n-r) !} \quad$ for $0 \leq r \leq n$

Expansion of $\left(1+x^{4}\right)^{3}={ }^{3} C_{0} 1^{3}+{ }^{3} C_{1} 1^{2} x^{4}+{ }^{3} C_{2} 1^{1} x^{8}+{ }^{3} C_{3} x^{12}$

and the general term in expansion of $(1-x)^{12}={ }^{12} C_{r} 1^{(12-r)} x^{r}$

The coefficient of $x^{8}$ in the expansion of $\left(1+x^{4}\right)^{3}(1-x)^{12}$ :

$\begin{array}{l} =\left({ }^{3} C_{0} 1^{3} \times{ }^{12} C_{8} 1^{4}\right)+\left({ }^{3} C_{1} 1^{2} \times{ }^{12} C_{4} 1^{8}\right)+\left({ }^{3} C_{2} 1^{1} \times{ }^{12} C_{0} 1^{12}\right) \\ =\left(1 \times{ }^{12} C_{8}\right)+\left(3 \times{ }^{12} C_{4}\right)+\left(3 \times{ }^{12} C_{0}\right) \\ ={ }^{12} C_{8}+3^{12} C_{4}+3 \\ \because{ }^{12} C_{8}={ }^{12} C_{4} \\ \therefore{ }^{12} C_{8}+3^{12} C_{4}+3=4{ }^{12} C_{4}+3 \end{array}$

So, the coefficient of $x^{8}=3+4 \times{ }^{12} C_{4}$