If $\sqrt{2}\cos A=\cos B+{\cos }^{3}Band\sqrt{2}\sin A=\sin B-{\sin }^{3}B$, then $\sin (A-B)=$

$\sqrt{2}\cos A=\cos B+{\cos }^{3}B\to \left(i\right)$

$\sqrt{2}\sin A=\sin B-{\sin }^{3}B\to \left(ii\right)$

$\Rightarrow \sqrt{2}\sin A\cdot \cos B-\sqrt{2}\cos A\cdot \sin B=(\sin B-{\sin }^{3}B)\cos B-(\cos B+{\cos }^{3}B)\sin B$

$\Rightarrow \sqrt{2}\sin (A-B)=-\sin B\cdot \cos B$

$\Rightarrow \sin (A-B)=\frac{-1}{2\sqrt{2}}\sin 2B$

Now squaring and adding equations (i) and (ii), we get

$2={\cos }^{2}B+{\sin }^{2}B+{\cos }^{6}B+{\sin }^{6}B+2({\cos }^{4}B-{\sin }^{4}B)$

$=1+(1-3{\cos }^{2}B{\sin }^{2}B)+2\cos 2B$

$2=2-\frac{3}{4}{\sin }^{2}2B+2\cos 2B$

$\Rightarrow 3{\cos }^{2}2B+8\cos 2B-3=0$

$\Rightarrow \cos 2B=\frac{1}{3}\Rightarrow \sin 2B=\pm \frac{2\sqrt{2}}{3}$

$\therefore \sin (A-B)=\pm \frac{1}{3}$.