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

\begin{align}
\sqrt 2 \sin A \cos B = \sin B \cos B \left ( 1-\sin^2B\right ) \tag 1 \\
\sqrt 2 \cos A \sin B = \sin B \cos B \left( 1 + \cos^2 B\right ) \tag 2
\end{align}
Now $(1) - (2)$ and use $\sin(A-B) = \sin A \cos B - \cos A \sin B$
\begin{align}
\sqrt 2  \sin (A-B) &= \sin B \cos B \left( 1-\sin^2B - 1 - \cos^2B\right ) = -\sin B \cos B \\
\sin(A-B) &= -\frac {\sin B\cos B}{\sqrt 2} \tag 3
\end{align}
\begin{align}
\sqrt 2 \cos A \cos B &= \cos^2B \left( 1+\cos^2 B\right ) \tag 4\\
\sqrt 2 \sin A \sin B &= \sin^2B \left( 1-\sin^2B\right ) = \sin^2 B \cos^2B\tag 5
 \end{align}
Now, $(4) + (5)$ and use $\cos(A-B) = \cos A \cos B - \sin A \sin B$
\begin{align}
\sqrt 2 \cos(A-B) &= \cos^2B \left( 1+\cos^2B + \sin^2B\right ) = 2\cos^2B \\
\cos(A-B) &= \sqrt 2 \cos^2B \tag 6
\end{align}
Now use $(3)^2+(6)^2 \equiv 1$, so
$$
\frac {\sin^2B \cos^2B}2 + 2 \cos^4B = 1 \\
\cos^2B \left( \sin^2B + 4 \cos^2B\right ) = 2 \\
\cos^2B \left ( 1 + 3 \cos^2B\right ) = 2 \\
3\cos^4B + \cos^2B - 2 = 0 \\
\cos^2B = \frac {-1 \pm \sqrt{1 + 24}}6 = \frac {-1 \pm 5}6 = \frac 23
$$
Therefore
$$
\sin^2B = 1 - \cos^2B = \frac 13
$$
and
$$
\sin B \cos B = \pm \sqrt{\sin^ 2B \cos^2B} = \pm \frac {\sqrt 2}3
$$
Finally
$$
\sin(A-B) = -\frac 1{\sqrt2} \sin B \cos B = \mp \frac 13
$$
