If ${\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)}^4={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\dots \dots \dots +{\mathrm{a}}_8{\mathrm{x}}^8$ where $\mathrm{a},\mathrm{b},{\mathrm{a}}_0,{\mathrm{a}}_1,{\mathrm{a}}_2\dots \dots {\mathrm{a}}_8\in \mathrm{R}$ such that ${\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2\ne 0\text{ and }\left|\begin{array}{l}{\mathrm{a}}_0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_2\\ {\mathrm{a}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_1\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_0\\ {\mathrm{a}}_2\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_0\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_1\end{array}\right|=0$ then 

$\mathrm{x}=0 : {\mathrm{a}}_0=1$ Differentiations

$4{\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)}^3(\mathrm{a}+2\mathrm{bx})={\mathrm{a}}_1+2{\mathrm{a}}_2\mathrm{x}+\dots$

            Let $\mathrm{x}=04\mathrm{a}={\mathrm{a}}_1$

The value of the determinant is

$-\left({\mathrm{a}}_0^3+{\mathrm{a}}_1^3+{\mathrm{a}}_2^3-3{\mathrm{a}}_0{\mathrm{a}}_1{\mathrm{a}}_2\right)=0$

$\Rightarrow \because {\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2\ne 0\text{ the }{\mathrm{a}}_0={\mathrm{a}}_1={\mathrm{a}}_2=4\mathrm{a}$

Differentially again and putting $\mathrm{x}=0$

$\mathrm{a}=\frac{1}{4} : \mathrm{b}=\frac{5}{32}$