#### Answer

$2\cos 6x\cdot\sin 3x$

#### Work Step by Step

Sum-to-Product:
$\displaystyle \sin A-\sin B=2\cos(\frac{A+B}{2})\sin(\frac{A-B}{2})$
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$\displaystyle \frac{A+B}{2}=\frac{9x+3x}{2}=6x$
$\displaystyle \frac{A-B}{2}=\frac{9x-3x}{2}=3x$
$\sin 9x-\sin 3x=2\cos 6x\cdot\sin 3x$