Nothing easier than this ........

Let $(a-b)=x$ , $(b-c)=y$ and $(c-a)=z$ .

$\dfrac{x^3+y^3+z^3}{xyz}$

$\dfrac{(x+y+z)(x^2+y^2+z^2-(xy+yz+zx))+3xyz}{xyz}$

$\dfrac{(a-b+b-c+c-a)(x^2+y^2+z^2-(xy+yz+zx)+3xyz}{xyz}$

$\dfrac{3xyz}{xyz}$

$\boxed{3}$

We know $x^3 + y^3 + z^3 = (x+y+z)^3 - 3(x+y+z)(xy+yz+xz) + 3xyz$

$\therefore (a - b)^3 + (b - c)^3 + (c - a)^3 = 0^3 - 3 \times 0 \times ( (a - b) (b - c)+(b - c) (c - a)+(a - b) (c - a) ) + 3(a - b) (b - c) (c - a)$

$(a - b)^3 + (b - c)^3 + (c - a)^3 = 3 (a - b) (b - c) (c - a)$

$\frac {(a - b)^3 + (b - c)^3 + (c - a)^3 }{(a - b) (b - c) (c - a)} = \boxed{3}$

Expanding the equation, we get

$\frac{a^{3} - b^{3} - 3a b(a-b) + b^{3} - c^{3} - 3bc(b-c) + c^{3} - c^{3} - 3ca(c-a)}{(abc - c^{2}a - b^{2}c + bc^{2} - a^{2}b + ca^{2} + ab^{2} - abc)}$

cancelling out

$\frac{3( -a^{2}b + ab^{2} - b^{2}c + bc^{2} - c^{2}a + ca^{2})}{(-a^{2}b + ab^{2} - b^{2}c + bc^{2} - c^{2}a + ca^{2})}$

cancelling out and answer is 3