$\sin (A-B) = \pm\frac{1}{3}.$ if....... please help me, how can I approach. I can not understand what to do. How can I start. Help me. If $\sqrt{2}\cos A =\cos B +\cos^3 B$ and $\sqrt{2} \sin A = \sin B - \sin^3 B$ show that $\sin (A-B) =\pm\frac{1}{3}$.

Well, to start with, we can square both equations to get $$2\cos^2A=\cos^2B+2\cos^4B+\cos^6B\\2\sin^2A=\sin^2B-2\sin^4B+\sin^6B$$ Adding these together and doing some factoring yields $$2(\sin^2A+\cos^2A)=\sin^2B+\cos^2B+2(\cos^2B-\sin^2B)(\sin^2B+\cos^2B)+\cos^6B+(\sin^2B)^3,$$ and applying the Pythagorean identity $\sin^2\theta=1-\cos^2\theta$, we find that $$2=1+2(2\cos^2B-1)+\cos^6B+(1-\cos^2B)^3\\2=1+4\cos^2B-2+\cos^6B+1-3\cos^2B+3\cos^4B-\cos^6B\\2=\cos^2B+3\cos^4B$$
Making the substitution $u=\cos^2B$ we have that $$3u^2+u-2=0\\3u^2+3u-2u-2=0\\3u(u+1)-2(u+1)=0\\(3u-2)(u+1)=0$$ so $u=\frac23$ or $u=-1,$ meaning $\cos^2B=-1$ (impossible for real $B$) or $\cos^2B=\frac23.$ Thus, we have $\cos B=\pm\sqrt{\frac23}.$ Let's address these cases separately.

Suppose that $\cos B=\sqrt{\frac23}.$ Applying the Pythagorean identity mentioned above once more, we find that $\sin^B=1-\cos^B=1-\frac23,$ and so $\sin B=\pm\sqrt{\frac13}.$ Then $$\sqrt2\cos A=\sqrt{\frac23}+\frac23\sqrt{\frac23}=\frac53\sqrt{\frac23},$$ and so $$\cos A=\frac53\sqrt{\frac13}.$$ Also, $$\sqrt2\sin A=\pm\sqrt{\frac13}\mp\frac13\sqrt{\frac13}=\pm\frac23\sqrt{\frac13},$$ and so $$\sin A=\pm\frac13\sqrt{\frac23}.$$ (The $\pm$ above corresponds exactly to the $\pm$ of $\sin B.$) Thus, using the identity $$\sin(A-B)=\sin A\cos B-\cos A\sin B,$$ we have $$\sin(A-B)=\left(\pm\frac13\sqrt{\frac23}\right)\left(\sqrt{\frac23}\right)-\left(\frac53\sqrt{\frac13}\right)\left(\pm\sqrt{\frac13}\right)=\pm\frac29\mp\frac59=\mp\frac39=\mp\frac13,$$ as desired.
The $\cos B=-\sqrt{\frac13}$ case is similar.
