$ {\left( {\sin x - \cos x} \right)^3} = 1 + \sin x\cos x\\  \Leftrightarrow {\sin ^3}x - 3\sin x\cos x\left( {\sin x - \cos x} \right) - {\cos ^3}x = 1 + \sin x\cos x\\  \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x + \sin x\cos x} \right) - 3\sin x\cos x\left( {\sin x - \cos x} \right) = 1 + \sin x\cos x\\  \Leftrightarrow \left( {\sin x - \cos x} \right)\left( {1 + \sin x\cos x} \right) - 3\sin x\cos x\left( {\sin x - \cos x} \right) - \left( {1 + \sin x\cos x} \right) = 0\\  \Leftrightarrow \left( {1 + \sin x\cos x} \right)\left( {\sin x - \cos x - 1} \right) - 3\sin x\cos x\left( {\sin x - \cos x} \right) = 0\\ t = \sin x - \cos x\left( {\left| t \right| \le \sqrt 2 } \right) \Rightarrow {t^2} = 1 - 2\sin x\cos x \Rightarrow \sin x\cos x = \dfrac{{1 - {t^2}}}{2}\\ PT \Leftrightarrow \left( {1 + \dfrac{{1 - {t^2}}}{2}} \right)\left( {t - 1} \right) - 3.\dfrac{{1 - {t^2}}}{2}t = 0\\  \Leftrightarrow \dfrac{{3 - {t^2}}}{2}\left( {t - 1} \right) - 3.\dfrac{{\left( {1 - {t^2}} \right)t}}{2} = 0\\  \Leftrightarrow \left( {3 - {t^2}} \right)\left( {t - 1} \right) - 3\left( {1 - t} \right)\left( {1 + t} \right)t = 0\\  \Leftrightarrow \left( {3 - {t^2}} \right)\left( {t - 1} \right) + 3\left( {t - 1} \right)\left( {t + 1} \right)t = 0\\  \Leftrightarrow \left( {t - 1} \right)\left[ {\left( {3 - {t^2}} \right) + 3\left( {t + 1} \right)t} \right] = 0\\  \Leftrightarrow \left( {t - 1} \right)\left( {2{t^2} + 3t + 3} \right) = 0\\  \Leftrightarrow t = 1 \Leftrightarrow \sin x - \cos x = 0 \Leftrightarrow \sqrt 2 \sin \left( {x - \dfrac{\pi }{4}} \right) = 0 \Rightarrow x - \dfrac{\pi }{4} = k\pi  \Rightarrow x = \dfrac{\pi }{4} + k\pi \left( {k \in Z} \right)\\ b){\sin ^3}x + {\cos ^3}x = 1\\  \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {{{\sin }^2}x + {{\cos }^2}x - \sin x\cos x} \right) = 1\\  \Leftrightarrow \left( {\sin x + \cos x} \right)\left( {1 - \sin x\cos x} \right) = 1\\ t = \sin x + \cos x\left( { - \sqrt 2  \le t \le \sqrt 2 } \right) \Rightarrow \sin x\cos x = \dfrac{{{t^2} - 1}}{2}\\ PT \Leftrightarrow t\left( {1 - \dfrac{{{t^2} - 1}}{2}} \right) = 1 \Leftrightarrow t\left( {\dfrac{{2 - {t^2} + 1}}{2}} \right) = 1 \Leftrightarrow t\left( {3 - {t^2}} \right) = 2 \Leftrightarrow {t^3} - 3t + 2 = 0\\  \Leftrightarrow {t^3} - {t^2} + {t^2} - t - 2t + 2 = 0\\  \Leftrightarrow \left( {t - 1} \right)\left( {{t^2} + t - 2} \right) = 0\\  \Leftrightarrow \left[ \begin{array}{l} t = 1\\ t =  - 2 \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) = 1\\ \sqrt 2 \sin \left( {x + \dfrac{\pi }{4}} \right) =  - 2(L) \end{array} \right. \Leftrightarrow \sin \left( {x + \dfrac{\pi }{4}} \right) = \dfrac{{\sqrt 2 }}{2}\\  \Leftrightarrow \left[ \begin{array}{l} x + \dfrac{\pi }{4} = \dfrac{\pi }{4} + k2\pi \\ x + \dfrac{\pi }{4} = \pi  - \dfrac{\pi }{4} + k2\pi  \end{array} \right. \Leftrightarrow \left[ \begin{array}{l} x = k2\pi \\ x = \dfrac{\pi }{2} + k2\pi  \end{array} \right.\left( {k \in Z} \right) $

`a)` `(sinx-cosx)^3=1+sinx cosx` $(1)$

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Đặt `t=sinx-cosx`

`=\sqrt{2}.(\sqrt{2}/2 sinx-\sqrt{2}/2cosx)`

`=\sqrt{2}.(cos\ π/4sinx-sin\ π/4 cosx)`

`=\sqrt{2}sin(x-π/4)`

Vì `|sin(x-π/4)|\le 1`

`=>| t|\le \sqrt{2}`

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`\qquad t^2=(sinx-cosx)^2`

`=>t^2=sin^2x+cos^2x-2sinx cosx`

`=>t^2=1-2sinx cosx`

`=>sinx cosx ={1-t^2}/2`

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`(1)<=>t^3=1+{1-t^2}/2`

`<=> t^3={2+1-t^2}/2`

`<=>2t^3=3-t^2<=>2t^3+t^2-3=0`

`<=>2t^3-2t^2+3t^2-3=0`

`<=>2t^2(t-1)+3(t+1)(t-1)=0`

`<=>(t-1)(2t^2+3t+3)=0`

`<=>`$\left[\begin{array}{l}t=1\\2t^2+3t+3=0\ (vô\ nghiệm)\end{array}\right.$

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`t=1<=>\sqrt{2}sin(x-π/4)=1`

`<=>sin(x-π/4)=1/\sqrt{2}=sin \ π/4`

`<=>`$\left[\begin{array}{l}x-\dfrac{π}{4}=\dfrac{π}{4}+k2π\\x-\dfrac{π}{4}=π-\dfrac{π}{4}+k2π\end{array}\right.$`(k\in ZZ)`

`<=>`$\left[\begin{array}{l}x=\dfrac{π}{2}+k2π\\x=π+k2π\end{array}\right.$`(k\in ZZ)`

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Vậy phương trình đã cho có họ nghiệm:

`\qquad x=π/2+k2π; x=π+k2π\ (k\in ZZ)`

`b)` `sin^3x+cos^3x=1`

`<=>(sinx+cosx)(sin^2 x-sinx cosx+cos^2x)=1`

`<=>(sinx+cosx)(1-sinx cosx)=1` `(**)`

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Đặt `t=sinx+cosx`

`=\sqrt{2}.(\sqrt{2}/2 sinx+\sqrt{2}/2cosx)`

`=\sqrt{2}.(cos\ π/4sinx+sin\ π/4 cosx)`

`=\sqrt{2}sin(x+π/4)`

Vì `|sin(x+π/4)|\le 1``=>| t|\le \sqrt{2}`

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`\qquad t^2=(sinx+cosx)^2`

`=>t^2=sin^2x+cos^2x+2sinx cosx`

`=>t^2=1+2sinx cosx`

`=>sinx cosx ={t^2-1}/2`

$\\$

`(**)<=>t.(1-{t^2-1}/2)=1`

`<=>t. {2-t^2+1}/2=1<=>t(3-t^2)=2`

`<=>-t^3+3t-2=0<=>`$\left[\begin{array}{l}t=-2\ (loại)\\t=1\ (thỏa\ mãn)\end{array}\right.$

Với `t=1`

`<=>\sqrt{2}sin(x+π/4)=1`

`<=>sin(x+π/4)=1/\sqrt{2}=sin \ π/4`

`<=>`$\left[\begin{array}{l}x+\dfrac{π}{4}=\dfrac{π}{4}+k2π\\x+\dfrac{π}{4}=π-\dfrac{π}{4}+k2π\end{array}\right.$`(k\in ZZ)`

`<=>`$\left[\begin{array}{l}x=k2π\\x=\dfrac{π}{2}+k2π\end{array}\right.$`(k\in ZZ)`

Vậy phương trình đã cho có họ nghiệm:

`\qquad x=k2π; x=π/2+k2π\ (k\in ZZ)`