In any triangle `A B C ,` prove that: `a^3cos(B-C)+b^3cos(C-A)+c^3cos(A-B)=3a b c`

Explanation

From sine law, we have,

`a/sinA = b/sinB = c/sinC = k`

`=> a = ksinA, b = ksin B, c = ksinC`

Now,

`L.H.S. = a^3cos(B-C)+b^3os(C-A)+c^3cos(A-B)`

`=a^2(ksinAcos(B-C))+b^2(ksinBcos(C-A))+c^2(ksinCcos(A-B))`

`=a^2(ksin(pi-(B+C))cos(B-C))+b^2(ksin(pi-(C+A))cos(C-A))+c^2(ksin(pi-(A+B))cos(A-B))`

`=k/2[a^2(2sin(B+C)cos(B-C))+b^2(2sin(C+A)sin(C-A))+c^2(2sin(A+B)cos(A-B))]`

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