If ${\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)}^4={\mathrm{a}}_0+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_2{\mathrm{x}}^2+\cdots +{\mathrm{a}}_8{\mathrm{x}}^8$, where $\mathrm{a}, \mathrm{b}, {\mathrm{a}}_0, {\mathrm{a}}_1, \dots ,{\mathrm{a}}_8\in \mathrm{R}$ such that ${\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2\ne 0$ and $\left|\begin{array}{ccc}{\mathrm{a}}_0& {\mathrm{a}}_1& \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_2\\ {\mathrm{a}}_1& {\mathrm{a}}_2& {\mathrm{a}}_0\\ {\mathrm{a}}_2& {\mathrm{a}}_0& {\mathrm{a}}_1\end{array}\right|=0$, then the value of $5\frac{\mathrm{a}}{\mathrm{b}}$ is ______ .

Putting x = 0, a₀ = 1

${\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)}^4=\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)\left(1+\mathrm{ax}+{\mathrm{bx}}^2\right)$

Clearly a₀ = 1, a₁ = coefficient of x = a + a + a + a = 4a

a₂ = coefficient of x² = 4b + 6a²

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

$\begin{array}{l}\because \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_0+{\mathrm{a}}_1+{\mathrm{a}}_2\ne 0\\ \therefore \hspace{0.25em}\hspace{0.25em}\hspace{0.25em}\hspace{0.25em}{\mathrm{a}}_0={\mathrm{a}}_1={\mathrm{a}}_2\end{array}$

1 = 4a = 6a² + 4b

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