$\sin^3x + \cos^3x = (\sin x + \cos x)(1 - \sin x \cos x ) = 1 - \sin^2 x \cos^2 x$ , temos: $(\sin x + \cos x)(1 - \sin x \cos x ) = (1 + \sin x \cos x )(1 - \sin x \cos x )$ Para $(1 - \sin x \cos x ) = 0$ $\implies$ $\frac{\sin(2x)}{2} = 1$ $\implies$ $\sin(2x) = 2$ (Não há soluções) Para $(1 - \sin x \cos x ) \neq 0$ $\implies$ $\sin x + \cos x = 1 + \sin x \cos x $ , temos: $(\sin x + \cos x)^2 = (1 + \sin x \cos x)^2 $ $\implies$ $\implies$ $1 + 2\sin x \cos x = 1 + 2\sin x \cos x + (\sin x \cos x)^2$ , portanto: $\sin x \cos x = \Large{\frac{\sin{2x}}{2}}$ $=$ $0$ $\implies$ $\sin(2x) = 0$ . Assim: $2x = 4k\pi$ $\implies$ $x = 2k\pi$ , $k \in \mathbb{Z}$ . Ou: $2x = \pi + 4k\pi$ $\implies$ $x = \frac{\pi}{2} + 2k\pi$ , $k \in \mathbb{Z}$ . $x = \{x \in \mathbb{R} \space | \space x = 2k\pi \space \space \text{ou} \space \space x = \frac{\pi}{2} + 2k\pi, k \in \mathbb{Z}\}$ $*$ Na equação original, quando $\sin(x) = 0$, então $\cos(x) = 1$ , ou seja, o $x$ é da forma $2k\pi$ .