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$\sin x-\cos x$ = m
$\Rightarrow $ $m^{2} = (\sin x)^{2} + (\cos x)^{2} - 2 \sin x\cos x$ = 1 - 2$\sin x\cos x$
$\Rightarrow \sin x\cos x = \frac{1-m^{2}}{2}$
$(\sin x)^{3}-(\cos x)^{3}$
=$(\sin x-\cos x)[(\sin x)^{2}+\sin x\cos x +(\cos x)^{2}]$
=$(\sin x-\cos x)[(\sin x-\cos x)^{2}+3\sin x\cos x)$
=$m[m^{2} + \frac{3}{2}(1-m^{2})]$
=$m(\frac{3-m^{2}}{2})$
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