1. Let $\sin x+\cos x=\frac{1}{2}$. Then $\sin ^{3} x+\cos ^{3} x=$ $\qquad$ .
$\begin{array}{l}\text { 1. } \frac{11}{16} \text {. } \\ \text { Given } \sin x+\cos x=\frac{1}{2} \\ \Rightarrow 1+2 \sin x \cdot \cos x=\frac{1}{4} \\ \Rightarrow \sin x \cdot \cos x=-\frac{3}{8} \\ \Rightarrow \sin ^{3} x+\cos ^{3} x \\ =(\sin x+\cos x)\left(\sin ^{2} x-\sin x \cdot \cos x+\cos ^{2} x\right) \\ =\frac{1}{2}\left(1+\frac{3}{8}\right)=\frac{11}{16} .\end{array}$