Do tan$\alpha$ = $\sqrt{2}$ > 0 nên $\alpha$ là góc nhọn và cos$\alpha$ > 0
K = $\frac{sin^{3}\alpha + sin\alpha . cos^{2}\alpha + 2sin^{2}\alpha . cos\alpha - 4cos^{3}\alpha}{sin\alpha - cos\alpha}$
K = $\frac{\frac{sin^{3}\alpha}{cos^{3}\alpha} + \frac{sin\alpha . cos^{2}\alpha}{cos^{3}\alpha} + \frac{2sin^{2}\alpha . cos\alpha}{cos^{3}\alpha} - \frac{4cos^{3}\alpha}{cos^{3}\alpha}}{\frac{sin\alpha}{cos^{3}\alpha} - \frac{cos\alpha}{cos^{3}\alpha}}$
K = $\frac{tan^{3}\alpha + tan\alpha + 2tan^{2}\alpha - 4}{tan\alpha.(1 + tan^{2}\alpha) - (1+tan^{2}\alpha)}$
K = $\frac{2\sqrt{2} + \sqrt{2} + 4 - 4}{\sqrt{2} . 3 - 3}$
K = $\sqrt{2}(\sqrt{2}+1)$