If $\lim _{x \rightarrow 0}\left(\cos x+a^{3} \sin \left(b^{6} x\right)\right)^{\frac{1}{x}}=e^{512}$ then value of $ab^2$ is equal to-

$\begin{array}{l}\lim _{x \rightarrow 0}\left[\cos x+a^{3} \sin \left(b^{6} x\right)\right]^{\frac{1}{x}} \\ =e^{\lim _{x \rightarrow 0} \frac{1}{x}\left[\cos x+a^{3} \sin \left(b^{6} x\right)-1\right]} \quad\left(1^{\infty \text { model }}\right)\end{array}$

$\begin{array}{l}=e^{\lim _{x \rightarrow 0}\left[-\left(\frac{1-\cos x}{x}\right)+\frac{a^{3} \sin \left(b^{6} x\right)}{x}\right]} \\ =e^{\lim _{x \rightarrow 0}\left[-\frac{2 \sin ^{2}\left(\frac{x}{2}\right)}{x}+\frac{a^{3} \sin \left(b^{6} x\right)}{x}\right]} \\ =e^{0+a^{3} b^{6}} \\ =e^{512} \quad \text { (given) } \\ \therefore a^{3} b^{6}=512\end{array}$

$\begin{array}{l}\Rightarrow\left(a b^{2}\right)^{3}=(8)^{3} \\ \therefore a b^{2}=8\end{array}$