[tex]\sin x +\cos x = \dfrac{1}{2}\Big|^2 \\ \\ \sin^2x+\cos^2 x+2\sin x\cos x = \dfrac{1}{4} \\ \\ 2\sin x\cos x = \dfrac{1}{4}-1\Rightarrow \sin x\cos x = -\dfrac{3}{8} \\ \\ \\ \sin^3 x+\cos^3 x = (\sin x+\cos x)(\sin^2 x-\sin x\cos x+\cos^2 x)\\ \\ = \dfrac{1}{2}\cdot (1+ \dfrac{3}{8}) = \dfrac{11}{16}[/tex]
