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

As Kaster has identified,
$(6)\displaystyle\sqrt2\cos(A-B)=2\cos^2B=\cos^2B-\sin^2B+1$
$\displaystyle\implies\cos^2B-\sin^2B=\sqrt2\cos(A-B)-1$
$(3)\displaystyle\cos B\sin B=-\sqrt2\sin(A-B)$
We have to eliminate $B$s keeping $A-B$
Using $\displaystyle(\cos^2B-\sin^2B)^2+4\cos^2B\sin^2B=(\cos^2B+\sin^2B)^2=1,$
$\displaystyle\{\sqrt2\cos(A-B)-1\}^2+4\{-\sqrt2\sin(A-B)\}^2=1$
$\displaystyle\implies 2\cos^2(A-B)-2\sqrt2\cos(A-B)+1+8\sin^2(A-B)=1$
Using $\displaystyle\sin^2(A-B)=1-\cos^2(A-B),$
$\displaystyle\implies -6\cos^2(A-B)-2\sqrt2\cos(A-B)+8=0$
Solve the Quadratic Eqaution for $\displaystyle\cos(A-B)$ and find that only one value lies in $\in[-1,1]$
Use $\displaystyle\sin(A-B)=\pm\sqrt{1-\cos^2(A-B)}$
