wht to do?

2\left(\frac{1}{1!.9!}+\frac{1}{3!.7!} \right)+\frac{1}{5!5!} \\\\=\frac{1}{10!}\left( \begin{pmatrix} 10\\1 \end{pmatrix}+\begin{pmatrix} 10\\3 \end{pmatrix}+\begin{pmatrix} 10\\5 \end{pmatrix}+\begin{pmatrix} 10\\7 \end{pmatrix}+\begin{pmatrix} 10\\9 \end{pmatrix}\right) \\\\=\frac{512}{10!}

Now,
x^3-5x^2+33x-19 \\=(x-2)^3+(x-2)^2+25(x-2)+35 \\=-125i-25+125i+35 \\=10

B is the answer.

Another method:

http://www.goiit.com/posts/list/algebra-in-complex-numbers-if-x-2-5i-then-the-value-of-x-3-5x-990384.htm#1194717