Монгол Бодлогын Сан

Заавар: Хуваарийг $\sin^2\alpha+\cos^2\alpha=1$ тоогоор үржүүл.

Бодолт: $$\dfrac{\sin^3\alpha+\cos^3\alpha}{3\cos\alpha-\sin\alpha}=\dfrac{\sin^3\alpha+\cos^3\alpha}{(3\cos\alpha-\sin\alpha)(\sin^2\alpha+\cos^3\alpha)}=$$ $$=\dfrac{\sin^3\alpha+\cos^3\alpha}{-\sin^3\alpha+3\sin^2\alpha\cos\alpha-\sin\alpha\cos^3\alpha+3\cos^2\alpha}=$$ $$=\dfrac{\tg^3\alpha+1}{-\tg^3\alpha+3\tg^2\alpha-\tg\alpha+3}=\dfrac{(-2)^3+1}{-(-2)^3+3\cdot(-2)^2-(-2)+3}=-\dfrac{7}{25}$$