Монгол Бодлогын Сан

Заавар: $a^6+b^2=(a^2+b^2)(a^4-a^2b^2+b^4)$ байна.

Бодолт: $\sin^6\alpha+\cos^6\alpha=(\sin^2\alpha+\cos^2\alpha)(\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha)$ ба $\sin^2\alpha+\cos^2\alpha=1$ тул $$\begin{align*} \sin^6\alpha&+\cos^6\alpha+3\sin^2\alpha\cos^2\alpha=\\ &=(\sin^4\alpha-\sin^2\alpha\cos^2\alpha+\cos^4\alpha)+3\sin^2\alpha\cos^2\alpha\\ &=\sin^4\alpha+2\sin^2\alpha\cos^2\alpha+\cos^4\alpha\\ &=(\sin^2\alpha)^2+2\sin^2\alpha\cos^2\alpha+(\cos^2\alpha)^2\\ &=(\sin^2\alpha+\cos^2\alpha)^2=1^2=1 \end{align*}$$