Prove that in `triangleABC, a^3cos(B-C)+b^3cos(C-A)+c^3cos(A-B)=3abc`

Explanation

`a^(3)cos(B-C)=a^(2).a.cos(B-C)`
`=a^(2)k.sin{(180^(@)-(B+C)}.cos(B-C)`
`=(a^(2)k)/(2)[sin2B+sin2C]`
`=(a^(2)k)/2[sin2B+sin2C]`
`=(a^(2)k)/(2)[2sinBcosB+2sinCcosC]`
`=a^(2)(ksinBcosB+ksinCcosC)`
`=a^(2)(bcosB+(c)cosC)`
Similarly, we can prove that
`b^(3)cos(C-A)=b^(2)(c)cosC+acosA`
and `c^(3)cos(A-B)=c^(2)(acosA+bcosB)`
Now L.H.S. `=a^(3)cos(B-C)+b^(3)cos(C-A)+c^(3)cos(A-B)`
`=a^(2)(bcosB+(c)cosC)+b^(2)((c)cosC+acosA) + c^(2)(a cosA+bcos B)`
`=a^(2)bcosB+a^(2)(c)cosC+b^(2)ccosC + b^(2)acosA+c^(2)acosA+c^(2)acosA+c^(2)bcosB`
`=(a^(2)bcosB+b^(2)acosA)+(a^(2)(c)cosC+c^(2)acosA)+(b^(2)(c)cosC+c^(2)bcosB)`
`ab(a cosB+bcosA)+ac(a cosC + (c)cosA) +bc(bcosC+c cosB)`
`=abc+acb+bca`
`=3abc`= RHS Hence Proved.