$sin^{3}x$ - $cos^{3}x$ = 1

\[\begin{array}{l} {\sin ^3}x - {\cos ^3} = 1 \Leftrightarrow {\sin ^3}x - \cos x\left( {1 - {{\sin }^2}x} \right) - 1 = 0\\ \Leftrightarrow {\sin ^3}x - \cos x + \cos x{\sin ^2}x - 1 = 0\\ \Leftrightarrow \left( {{{\sin }^3}x - 1} \right) + \cos x\left( {{{\sin }^2}x - 1} \right) = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {{{\sin }^2}x + \sin x + 1} \right) + \cos x\left( {\sin x - 1} \right)\left( {\sin x + 1} \right) = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left[ {{{\sin }^2}x + \sin x + 1 + \cos x\left( {\sin x + 1} \right)} \right] = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left[ {1 - {{\cos }^2}x + \left( {\sin x + 1} \right) + \cos x\left( {\sin x + 1} \right)} \right] = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left[ {\left( {1 - \cos x} \right)\left( {1 + \cos x} \right) + \left( {\sin x + 1} \right)\left( {1 + \cos x} \right)} \right] = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {1 + \cos x} \right)\left( {1 - \cos x + \sin x + 1} \right) = 0\\ \Leftrightarrow \left( {\sin x - 1} \right)\left( {1 + \cos x} \right)\left( {2 - \cos x + \sin x} \right) = 0\\ \Leftrightarrow \left[ \begin{array}{l} \sin x = 1\\ \cos x = - 1\\ \cos x - \sin x = 2\,\,\,\,\left( {VN} \right) \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = \frac{\pi }{2} + k2\pi \\ x = \pi + k2\pi \end{array} \right.\,\,\,\left( {k \in Z} \right). \end{array}\]