Quay lại đáp án

$\sin x-\cos x$ = 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})$

$\sin x-\cos x$ = 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})$