Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$

Notice $\boxed{\sin \alpha = 2\cos \alpha}$ and $$\cos ^2\alpha = {1\over 1+\tan^2\alpha} ={1\over 5}$$ so we have $$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}=\frac{8\cos^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{6\cos\alpha +2\cos\alpha}$$

$$= \frac{6\cos^{3}\alpha + 3\cos\alpha}{8\cos\alpha} = \frac{6\cos^{2}\alpha + 3}{8} = {21\over 40}$$

$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}=\frac{\tan^{3}\alpha - 2+ 3\sec^2\alpha}{(3\tan\alpha +2)\sec^2\alpha}$$

where $$\sec^2\alpha=\tan^2\alpha+1.$$

Hence $$\frac{21}{40}.$$

$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha} = \frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha} \cdot\frac{1/\cos^3\alpha}{1/\cos^3\alpha} = \frac{\tan^3\alpha-2+3\cdot(1/\cos^2\alpha)}{(3\tan\alpha+2)\cdot(1/\cos^2\alpha)}$$ Now, recall that $\frac{1}{\cos^2\alpha}=\sec^2\alpha=1+\tan^2\alpha=5$, so,

$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha} = \frac{\tan^3\alpha-2+3\cdot(1/\cos^2\alpha)}{(3\tan\alpha+2)\cdot(1/\cos^2\alpha)} = \frac{8-2+15}{(6+2)5}=\frac{21}{40}$$