`Answer:`

\(\sin^3\left(x-\frac{\pi}{6}\right)+3\sin^3\left(x+\frac{\pi}{3}\right)=\cos x+\sin2x\)

\(\Leftrightarrow\left[\sin\left(x-\frac{\pi}{6}\right)\right]^3+3\left[\sin\left(x+\frac{\pi}{3}\right)\right]^3=\cos x+\sin2x\)

\(\Leftrightarrow\left(\sin x.\cos\frac{\pi}{6}-\sin\frac{\pi}{6}\cos x\right)^3+3\left(\sin x\cos\frac{\pi}{3}+\sin\frac{\pi}{3}\cos x\right)^3=\cos x+\sin2x\)

\(\Leftrightarrow\left(\frac{\sqrt{3}}{2}\sin x-\frac{1}{2}\cos x\right)^3+3\left(\frac{1}{2}\sin x+\frac{\sqrt{3}}{2}\cos x\right)^3=\cos x+\sin2x\)

\(\Leftrightarrow\left(\sqrt{3}\sin x-\cos x\right)^3+3\left(\sin x+\sqrt{3}\cos x\right)^3=8\cos x+8\sin2x\)

\(\Leftrightarrow\left(\sqrt{3}\sin x\right)^3-3\left(\sqrt{3}\sin x\right)^2\cos x+3\sqrt{3}\sin x\cos^2x-\cos^3x+3\left(\sin^3x+3\sin^2x\sqrt{3}\cos x+3\sin x\left(\sqrt{3}\cos x\right)^2+\left(\sqrt{3}\cos x\right)^3\right)=8\cos x+8\sin2x\)

\(\Leftrightarrow3\sqrt{3}\sin^3x-9\sin^2x\cos x+3\sqrt{3}\sin x\cos^2x-\cos^3x+3\sin^3x+9\sqrt{3}\cos x\left(\sin^2x+\cos^2x\right)+27\sin x\cos^2x=8\cos x+8\sin2x\)

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