$\dfrac {a^3 + b^3 - c^3 + 3abc}{a^2 + b^2+c^2 - ab + bc + ca}=$
Solution
$a^3 + b^3 - c^3 + 3abc = a^3 + b^3 + (-c)^3 - 3ab (-c)$
$= (a + b - c) (a^2 + b^2 +c^2 -ab + bc + ca)$
$\therefore \dfrac{a^3 + b^3 - c^3 + 3abc} {a^2 + b^2 +c^2 - ab + bc +ca} = a + b - c$