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! :) 

Note that
$$\tan \alpha = \frac{\sin \alpha}{\cos \alpha}$$
Thus if
$$\tan \alpha = 2$$
then
$$\sin \alpha = 2 \cos \alpha$$
Now just plug for sine
$$\frac{\sin^3 \alpha - 2 \cos^3 \alpha + 3 \cos \alpha}{3 \sin \alpha + 2 \cos \alpha} = \frac{8 \cos^3 \alpha - 2 \cos^3 \alpha + 3 \cos \alpha}{6 \cos \alpha + 2 \cos \alpha}$$
which then simplifies to
$$\frac{6 \cos^3 \alpha + 3 \cos \alpha}{8 \cos \alpha} = \frac{1}{8} [6 \cos^2 \alpha + 3]$$
Now note that
$$\frac{1}{\cos \alpha} = \sec \alpha$$
and we have the trigonometric identity
$$1 + \tan^2 \alpha = \sec^2 \alpha$$
thus $$\sec^2 \alpha = 1 + (2)^2 = 1 + 4 = 5$$ and thus $\cos^2 \alpha = \frac{1}{5}$. Thus we can plug that into the prior expression to get
$$\frac{\sin^3 \alpha - 2 \cos^3 \alpha + 3 \cos \alpha}{3 \sin \alpha + 2 \cos \alpha} = \frac{1}{8} [6 \cos^2 \alpha + 3] = \frac{1}{8} [6 \frac{1}{5} + 3] = \frac{21}{40}$$.
