<s>[INST] Given $\tan\alpha=2$, evaluate $\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$ I need some help with this exercise. 

Given that $$\tan\alpha=2$$
  calculate the value of:
  $$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}$$

I've tried everything but I always end up having something irreducible in the end. Maybe this is an easy question, but I'm new at trigonometry. If anyone can help me by providing a step-by-step solution to this, I would be really thankful! :) 
 [/INST] Notice $\boxed{\sin \alpha = 2\cos \alpha}$ and  $$\cos ^2\alpha = {1\over 1+\tan^2\alpha} ={1\over 5}$$
so we have
$$\frac{\sin^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{3\sin\alpha +2\cos\alpha}=\frac{8\cos^{3}\alpha - 2\cos^{3}\alpha + 3\cos\alpha}{6\cos\alpha +2\cos\alpha}$$
$$= \frac{6\cos^{3}\alpha + 3\cos\alpha}{8\cos\alpha} = \frac{6\cos^{2}\alpha + 3}{8} = {21\over 40}$$
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