$\begin{array}{l}1)\,\,\sin^3x - \cos^3x = \sin x + \cos x\\ \Leftrightarrow \sin x(\sin^2x - 1) -\cos x(\cos^2x - 1)=0\\ \Leftrightarrow -\sin x\cos^2x + \cos x\sin^2x = 0\\ \Leftrightarrow \sin x\cos x(\sin x - \cos x) = 0\\ \Leftrightarrow \sin2x.\sin\left(x - \dfrac{\pi}{4}\right) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin2x = 0\\\sin\left(x - \dfrac{\pi}{4}\right) = 0\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}2x = k\pi\\x - \dfrac{\pi}{4} = k\pi\end{array}\right.\\ \Leftrightarrow \left[\begin{array}{l}x = k\dfrac{\pi}{2}\\x = \dfrac{\pi}{4} + k\pi\end{array}\right.\quad (k \in \Bbb Z)\\ 2\,\,4(\sin^3x + \cos^3x) = \sin x + \cos x\\ \Leftrightarrow 4(\sin x + \cos x)(1 - \sin x\cos x) - (\sin x + \cos x) = 0\\ \Leftrightarrow (\sin x + \cos x)(3 - 4\sin x\cos x) = 0\\ \Leftrightarrow \sin\left(x + \dfrac{\pi}{4}\right)(3 - 2\sin2x) = 0\\ \Leftrightarrow \left[\begin{array}{l}\sin\left(x + \dfrac{\pi}{4}\right) = 0\\\sin2x = \dfrac{3}{2}\quad (loại)\end{array}\right.\\ \Leftrightarrow x + \dfrac{\pi}{4} =k\pi\\ \Leftrightarrow x = - \dfrac{\pi}{4} +k\pi \quad (k \in \Bbb Z) \end{array}$
