Question

prove that: (x-a)³+(x-b)³+(x-c)³-3(x-a)(x-b)(x-c)=0, when a+b+c= 3x

Answer 1

this eq is in form of A^3+B^3+C^3-3ABC
A+B+C whole cube
but here A=x-a via
A+B+C wholecube =3x-a-b-c whole cube [a+b+c=3x;3x-a-b-c=0]
                                 =0 hence proved