Notice that
\(\dfrac { { a }^{ 3 }(b+c) }{ (a-b)(a-c) } +\dfrac { { b }^{ 3 }(a+c) }{ (b-a)(b-c) } +\dfrac { { c }^{ 3 }(b+a) }{ (c-b)(c-a) } = \dfrac{a^3(b + c)(c- b) + b^3(a + c)(a - c) + c^3(b + a)(b - a)}{(a - b)(b - c)(c - a)}\)
Expanding the numerator gives \(a^3(c^2 - b^2) + b^3 (a^2 - c^2) + c^3 (b^2- a^2)\).
Now, notice that the numerator is a homogeneous polynomial of degree 5. We let \(f(a, b, c) = a^3(c^2 - b^2) + b^3 (a^2 - c^2) + c^3 (b^2- a^2)\), which is the numerator.
We consider \(f(a, a, c) = a^3(c^2 - a^2) + a^3 (a^2 - c^2) + c^3 (a^2- a^2)\).
\(\quad f(a, a, c) \\ = a^3(c^2 - a^2) + a^3 (a^2 - c^2) + c^3 (a^2- a^2)\\ = a^3 (c^2 - a^2 + a^2 - c^2)\\ = 0\)
By factor theorem, (a - b) is a factor of f(a, b, c).
By properties of homogeneous polynomials, (b - c) and (c - a) are also factors of f(a, b, c).
Now, consider f(a, b, c) again, and we try to factor out these common factors we just found.
\(\quad f(a, b, c) \\= a^3(c^2 - b^2) + b^3 (a^2 - c^2) + c^3 (b^2- a^2)\\ = a^3 c^2 - a^3 b^2 + a^2 b^3 - b^3 c^2 + b^2 c^3 - a^2 c^3\\ = a^2 c^2 (a - c)+ b^3(a^2 - c^2) + b^2 (c^3 - a^3)\\ = a^2 c^2 (a - c) + b^3 (a + c)(a - c) - b^2 (a - c)(a^2 + ac + c^2)\\ = (a - c)(a^2 c^2 +b^3(a + c) - b^2(a^2 + ac +c^2))\\ = (a - c)(a^2 c^2 + ab^3 + b^3c - a^2b^2 - ab^2c - b^2 c^2)\\ = (a - c)(ab^2(b - c) + b^2c(b - c) + a^2 (c^2 - b^2))\\ = (a - c)(b - c)(ab^2 + b^2c - a^2(b + c))\\ = (a - c)(b - c)(ab(b - a)+ c(b^2 - a^2))\\ = (a - c)(b - c)(a - b)(-ab-c(a + b))\\ = (a - b)(b - c)(c - a)(ab + bc + ca)\)
Therefore, the original expression is \(\dfrac{f(a, b, c)}{(a - b)(b - c)(c - a)} = \dfrac{(a - b)(b - c)(c - a)(ab + bc + ca)}{(a - b)(b - c)(c - a)} = \boxed{ab + bc + ca}\)
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