Let $\mathrm{a}, \mathrm{b}\in \mathrm{R}$ and $\mathrm{ab}\ne 1$. If $6{\mathrm{a}}^2+20\mathrm{a}+15=0$ and $15{\mathrm{b}}^2+20\mathrm{b}+6=0$ then the value of $\frac{4030{\mathrm{b}}^3}{{\mathrm{ab}}^2-9(\mathrm{ab}+1{)}^3}$ is __________.

From the given relations, we can say that a and $\frac{1}{\mathrm{b}}$ are the roots of the equation 6x² + 20x + 15 = 0.

$\therefore \mathrm{a}+\frac{1}{\mathrm{b}}=-\frac{10}{3}\text{ and }\frac{\mathrm{a}}{\mathrm{b}}=\frac{5}{2}$

$\begin{array}{c}\frac{{\mathrm{b}}^3}{{\mathrm{ab}}^2-9(\mathrm{ab}+1{)}^3}=\frac{1}{\mathrm{a}\cdot \frac{1}{\mathrm{b}}-9{\left(\mathrm{a}+\frac{1}{\mathrm{b}}\right)}^3}\\ =\frac{1}{\frac{5}{2}+9\cdot \frac{1000}{27}}=\frac{6}{2015}\end{array}$