${\left(1+\mathrm{x}+{\mathrm{x}}^2\right)}^8={\mathrm{a}}_0+{\mathrm{a}}_1+\dots \dots +{\mathrm{a}}_{16}{\mathrm{x}}^{16}$ then ${\mathrm{a}}_1-{\mathrm{a}}_3+{\mathrm{a}}_5-{\mathrm{a}}_7+\dots \dots -{\mathrm{a}}_{15}=$

${(1+\mathrm{x}+{\mathrm{x}}^2)}^8={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+.....+{\mathrm{a}}_{16}{\mathrm{x}}^1 -\left(1\right)$

Put -x in place of x

${(1-\mathrm{x}+{\mathrm{x}}^2)}^2={\mathrm{a}}_0-{\mathrm{a}}_1 \mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2-{\mathrm{a}}_3{\mathrm{x}}^3+....+{\mathrm{a}}_{16}{\mathrm{x}}^{16} -\left(2\right)$

$$\begin{gathered} \left(1\right)-\left(2\right) \\ {(1+\mathrm{x}+{\mathrm{x}}^2)}^8-{(1-\mathrm{x}+{\mathrm{x}}^2)}^8 \end{gathered}$$

$\begin{array}{rcl} & =& 2\mathrm{x} \left[{\mathrm{a}}_1+{\mathrm{a}}_3{\mathrm{x}}^2+.....+{\mathrm{a}}_{15}{\mathrm{x}}^{14}\right]\\ \frac{{(1+\mathrm{x}+{\mathrm{x}}^2)}^8-{(1-\mathrm{x}+{\mathrm{x}}^2)}^8}{2\mathrm{x}}& =& {\mathrm{a}}_1+{\mathrm{a}}_3{\mathrm{x}}^2+.....+{\mathrm{a}}_{15}{\mathrm{x}}^{14}\\ \mathrm{Put} \mathrm{x}& =& \mathrm{i}\\ \frac{{(1+\mathrm{i}+{\mathrm{i}}^2)}^8-{(1-\mathrm{i}+{\mathrm{i}}^2)}^8}{2\mathrm{i}}& =& {\mathrm{a}}_1+{\mathrm{a}}_3 {\mathrm{i}}^2+....+{\mathrm{a}}_{15} {\mathrm{i}}^{14}\\ \frac{1-1}{2\mathrm{i}}& =& 0={\mathrm{a}}_1-{\mathrm{a}}_3+{\mathrm{a}}_5-{\mathrm{a}}_7+....+{\mathrm{a}}_{13}-{\mathrm{a}}_{15}\\ {\mathrm{a}}_1-{\mathrm{a}}_3+{\mathrm{a}}_5-{\mathrm{a}}_7+.....+{\mathrm{a}}_{13}-{\mathrm{a}}_{15}& =& 0\end{array}$