There's something special!

$ax^2+bx+c-\overline{abc} =0\\ ax^2 + bx + c - (100a+10b+c) = 0\\ ax^2 + bx - 100a - 10b = 0\\ x=\dfrac{-b \pm \sqrt{b^2 - 4a(-100a-10b)}}{2a}\\ =\dfrac{-b \pm \sqrt{400a^2 + 40ab + b^2}}{2a}\\ =\dfrac{-b \pm \sqrt{(20a)^2 + 2(20a)(b) + b^2}}{2a}\\ =\dfrac{-b \pm \sqrt{(20a+b)^2}}{2a}\\ =\dfrac{-b \pm (20a+b)}{2a}$

Now, let's try the solution with $+(20a+b)$

$x = \dfrac{-b + (20a+b)}{2a} = \dfrac{20a}{2a} = 10$

This solution is an integer, therefore $k=10$

$k^3 + 7k^2 + 2k + 9 = 10^3 + 7(10)^2 +2 (10) + 9 = \boxed{1729}$

The other solution, however, can also be an integer.

$x=\dfrac{-b - (20a+b)}{2a} = \dfrac{-2b - 20a}{2a} = -\dfrac{b}{a} - 10$

This solution will be an integer too if $|b|$ is a multiple of $|a|$

For example, $a=2,\;b=4$

$x = -\dfrac{4}{2} - 10 = -12$

In this case, $k$ can also be $-12$ .

Then, $k^3 + 7k^2 + 2k + 9 = (-12)^3 + 7(-12)^2 +2 (-12) + 9 = \boxed{-735}$

Of course, there are many other possible answers.

The only way to ensure that this question has a unique answer is to add the condition: $|b|$ is not a multiple of $|a|$