If $\tan \theta=2$, then the value of $\frac{8 \sin \theta+5 \cos \theta}{\sin ^{3} \theta+2 \cos ^{3} \theta+3 \cos \theta}$ is

Expression $=\frac{8 \sin \theta+5 \cos \theta}{\sin ^{3} \theta+2 \cos ^{2} \theta+3 \cos \theta}$

Dividing numerator and denominator by cos θ

$=\frac{8 \tan \theta+5}{\tan \theta \cdot \sin ^{2} \theta+2 \cos ^{2} \theta+3}$

$=\frac{8 \tan \theta+5}{2 \sin ^{2} \theta+2 \cos ^{2} \theta+3}=\frac{8 \tan \theta+5}{2\left(\sin ^{2} \theta+\cos ^{2} \theta\right)+3}$

$=\frac{8 \times 2+5}{5}=\frac{21}{5}$