10. If $\sin ^{3} x-\cos ^{3} x=-1$, then $\sin x-\cos x=$
10. Let $\sin x - \cos x = t \left(t \in [-\sqrt{2}, \sqrt{2}]\right)$, then $\sin x \cos x = \frac{1 - t^2}{3}$. From $\sin^3 x - \cos^3 x = -1$, we get $t \left(1 + \frac{1 - t^2}{3}\right) = -1$, which simplifies to $(t + 1)(t^2 - t - 3) = 0$, so $t = -1$.