Hi Gaurav,
We know the formula that,
$$A^3\ +\ B^3\ +\ C^3\ -\ 3ABC\ =\ \left(A\ +\ B\ +\ C\right)\left(A^2\ +\ B^2\ +\ C^2\ -\ AB\ -\ BC\ -\ CA\right)$$
So, when A + B + C = 0, the value of RHS becomes 0. As we multiply anything with 0, the answer is 0.
The value of LHS is $$A^3\ +\ B^3\ +\ C^3\ -\ 3ABC\ $$ which has to be 0 at A + B + C = 0.
$$A^3\ +\ B^3\ +\ C^3\ -\ 3ABC\ $$ = 0
$$A^3\ +\ B^3\ +\ C^3\ =\ 3ABC\ $$
In the question given let us assume that A = a - b, B = b - c and C = c - a,
Then the value of A + B + C = a - b + b - c + c - a = 0.
Hence, the value of $$A^3\ +\ B^3\ +\ C^3\ =\ 3ABC\ $$
$$\left(a\ -\ b\right)^3\ +\ \left(b\ -\ c\right)^3\ +\ \left(c\ -\ a\right)^3\ =\ 3\left(a\ -\ b\right)\left(b\ -\ c\right)\left(c\ -\ a\right)\ $$
Hope this helps!