7. Given $4 \sin \theta \cos \theta-5 \sin \theta-5 \cos \theta-1=0, \sin ^{3} \theta+\cos ^{3} \theta=$
7. $-\frac{11}{16}$.

Let $\sin \theta+\cos \theta=t$, then $t^{2}-1=2 \sin \theta \cos \theta,|t| \leqslant \sqrt{2} .2\left(t^{2}-1\right)-5 t-1=0,2 t^{2}-5 t-$ $3=0, t=3$ (discard) or $t=-\frac{1}{2}, \sin \theta+\cos \theta=-\frac{1}{2}, \sin \theta \cos \theta=-\frac{3}{8}$. And $\sin ^{3} \theta+\cos ^{3} \theta=$ $(\sin \theta+\cos \theta)(1-\sin \theta \cos \theta)=-\frac{11}{16}$.