Prove that, in triangle ABC `sin^3A cos(B-C)+sin^3B cos(C-A)+sin^3C cos(A-B)=3sinAsinBsinC`

Explanation

`LHS =sin^(2)A(B+C)cos(B-C)+sin^(2)Bsin(C+A)`
`cos(C-A)+sin^(2)Csin(A+B)cos(A-B)`
`=(1)/(2)sin^(2)A(sin 2B+sin2C)+(1)/(2)sin^(2)B`
`(Sin 2C+sin2A)+(1)/(2)sin^(2)C(sin 2A+sin2B)`
`=sin^(2)A(sin B cos B+sin Ccos C)`
`+sin^(2)B(sin C cos C+sin A cos A)`
`+sin^(2)C(sin A cos A +sin B cos B)`
`=sin A sin B (sin A cos B+cosB +cos A sin B)`
`+sin B sin C (sin B cos C+cos B sin C)`
`+sin C sin A (sin A cos C+cos A sin C)`
`=sin A + sin B sin (A+B)+sin B sin C sin (B+C)`
`+sin C sin A sin (A+C)`
`=3 sin A sin C=RHS`.