Integral roots of cubic equation $x^3-27x+k=0$ The number of integers $k$ for which the equation $x^3-27x+k=0$ has atleast two distinct integer roots is
(A)$1$ 
(B)$2$ 
(C)$3 $
(D)$4$
My Attempt: The condition for cubic $x^3+ax+b=0$to have $3$ real roots happens to be $4a^3+27b^2\leq0$. But how to go about finding condition for integer roots.

If $x=b$ is one of the solutions
$$k=27b-b^3$$
Now if $b$ is a repeated root and $c$ is the third one,
$0=b+b+c\iff c=-2b$
$$\implies x^3-27x+k=(x-b)^2(x+2b)=x^3+x(2b+b^2)-2b^3$$
$\implies b^2+2b=-27\iff b^2+2b+27=0$ which does not have an integer solution.
So, we can not have repeated roots.
The rest two solutions will be available from the quadratic equation $$0=\dfrac{x^3-27x-(b^3-27b)}{x-b}=x^2+bx+b^2-27$$
As $x$ is an integer, the he discriminant must be perfect square i.e.,  
$b^2-4(b^2-27)=108-3b^2=3(36-b^2)=D$(say)
$\implies36-b^2\ge0\iff b^2\le36\implies b\le6$
Also $3$ must divide $b$ to keep $D$ perfect square
So, $b\in[0,\pm3,\pm6]$
Clearly, $D$ is perfect square only for $b=\pm6$ .
