Let $a, b, c$ be the three roots of $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$ 
Let $a, b, c$ be the three roots of the equation $x^3-(k+1)x^2+kx+12=0$, where $k$ is a real number. If $(a-2)^3+(b-2)^3+(c-2)^3=-18$, find the value of $k$.

Here is a problem from a Hong Kong competition that I found a way to solve, but I am interested in knowing other methods since my brute-force way doesn't seem to be the solution intended.
My attempt
Let $y=x-2$, the equation becomes
$$(2+y)^3-(k+1)(2+y)^2+k(2+6)+12=0$$
$$y^3+(5-k)y^2+(8-3k)y-2k+16=0$$
Let $y_1, y_2, y_3$ be the new roots of the equation. By Vieta's formula, we have
\begin{cases}
y_1+y_2+y_3=-5+k \\
y_1y_2+y_2y_3+y_1y_3=8-3k\\
y_1y_2y_3=2k-16
\end{cases}
And by $y_1^3+y_2^3+y_3^3 = (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3)+3y_1y_2y_3$, we have
\begin{align*}
y_1^3+y_2^3+y_3^3 &= (y_1+y_2+y_3)^3-3(y_1+y_2+y_3)(y_1y_2+y_2y_3+y_1y_3) +3y_1y_2y_3\\
&= (-5+k)^3-3(-5+k)(8-3k)+3(2k-16)\\
&=k^3-6k^2+12k-53
\end{align*}
By $y_1^3+y_2^3+y_3^3=-18$ given in the problem, we have
$$k^3-6k^2+12k-35=0$$
And $k=5$ is the only real solution.
 A: Here is Stinking Bishop's method:
The condition on $a,b,c$ gives $(a^3 + b^3 + c^3) + (-2 \cdot 3)(a^2 + b^2 + c^2) + (4 \cdot 3)(a + b + c) - 3 \cdot 2^3 = -18$ or that $(a^3 + b^3 + c^3) - 6(a^2 + b^2 + c^2) + 12(a + b + c) = 6$.
Then as you said, $(a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc - 6((a+b+c)^2 - 2(ab+bc+ca))$ $+\ 12(a+b+c) = 6$, so by Vieta, $(k+1)^3 - 3(k+1)(k) + 3(-12) - 6((k+1)^2 - 2k) + 12(k+1) = 6$, or $k^3 + 3k^2 + 3k + 1 - 3k^2 -3k - 36 - 6k^2 - 12k - 6 + 12k + 12k + 12 = 6$, which implies $f(k) = k^3 - 6k^2 + 12k - 35 = 0$.
Descartes' rule of signs implies there are either three or one positive roots, and since $f(-k) = -k^3 - 6k^2 - 12k - 35 = 0$, there are no negative roots. Together with the rational root theorem, the only rational candidates (it is unlikely $k$ is irrational in a contest problem) are the factors of $35$: $35, 7, 5, 1$, and $5$ is a root. Dividing by $k = 5$ gives $k^2 - k + 7$ where $\Delta > 0$, hence $k = 5$ is the only solution.
A: using the identity $$a^3+b^3+c^3=3abc+\frac{(a+b+c)}{2} ({(a-b)}^2+{(b-c)}^2+{(c-a)}^2)$$ $$-18=3(a-2)(b-2)(c-2)+\frac{a+b+c-6}{2}({(a-b)}^2+{(b-c)}^2+{(c-a)}^2)$$ Now $$x^3-(k+1)x^2+kx+12=(x-a)(x-b)(x-c)$$ $$\to (a-2)(b-2)(c-2)=2k-16$$ now use $${(a-b)}^2+{(b-c)}^2+{(c-a)}^2={(a+b+c)}^2-3(ab+bc+ca)={(k+1)}^2-3k$$ combining these we get the cubic $$k^3-6k^2+12k-35=0$$
