<s>[INST] cubic equation has $3$ distinct roots $\alpha,\beta,\gamma$ equation $x^3-9x^2+24x+c=0$ has $3$ distinct roots $\alpha,\beta,\gamma$, then $\lfloor \alpha \rfloor+\lfloor \beta \rfloor +\lfloor \gamma \rfloor =$
and $\lfloor \alpha \rfloor = \alpha-\{\alpha\},\lfloor \beta \rfloor = \beta-\{\beta\},\lfloor \gamma \rfloor = \gamma-\{\gamma\},0\leq \{\alpha\},\{\beta\},\{\gamma\}<1$
we assume $f(x) = x^3-9x^2+24x+c$, using $f'(x) = 3x^2-18x+24$
put maximum , minimum  $f'(x)=0$
$3x^2-18x+24=0$
$x^2-6x+8=(x-4)(x-2)=0$
$x=2,x=4$
put $x=2,3$ in $f''(x)=6x-18=6(x-3)$ 
$f''(2) = -6<0$ means $x=2$ is a point of local maximum
and $f''(4)=6>0$ means $x=4$ is a point of local minimum
i can not go further
 [/INST] We may assume that $\alpha\lt \beta\lt \gamma$.
We have to have
$$f(2)=20+c\gt 0\quad\text{and}\quad f(4)=16+c\lt 0$$
Note that we have
$$f(1)=16+c\lt 0\quad\text{and}\quad f(5)=20+c\gt 0$$
from which
$$1\lt\alpha\lt 2\implies 0\lt\{\alpha\}\lt 1$$
$$4\lt\gamma\lt 5\implies 0\lt\{\gamma\}\lt 1$$
follow.
Thus, since
$$0\lt \{\alpha\}+\{\beta\}+\{\gamma\}\lt 3\implies \{\alpha\}+\{\beta\}+\{\gamma\}=1,2$$
and
$$\alpha+\beta+\gamma=9$$
we get
$$\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor=9-(\{\alpha\}+\{\beta\}+\{\gamma\})=\color{red}{7,8}$$
(Since $f(3)=18+c$, we have that $\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor=7$ for $-20\lt c\lt -18$ and that $\lfloor\alpha\rfloor+\lfloor\beta\rfloor+\lfloor\gamma\rfloor=8$ for $-18\le c\lt -16$.)
</s>