Rational roots of a cubic polynomial Find all distinct non-zero rational numbers $a$, $b$ and $c$ such that $x (x+a) (x+b) +c$ has 3 distinct non-zero rational roots.
What I have so far:
Let the polynomial have factorization $(x+x_1)(x+x_2)(x+x_3)$ where $x_i$'s are distinct non-zero
rational number. Equate coefficients of like power,
$$(x+x_1)(x+x_2)(x+x_3) = x (x+a) (x+b) +c,$$
after some algebraic manipulation we get:
$$
x_3^2  -  (a+b - x_1) x_3 + (a \cdot b - x_1 (a+b - x_1)) = 0
$$
Since $x_3$ is assumed to be rational, the discriminate must be rational
$$
r = \sqrt{(a+b - x_1)^2 - 4  (a \cdot b - x_1 (a+b - x_1))} 
$$
Assuming we can choose $a$, $b$ and $x_1$ such that $r$ is rational,
\begin{align*}
x_3 & = \frac{a + b - x_1 \pm r}{2} \\
x_2 &= a + b - x_1 - x_3 \\
c &= x_1 x_2 x_3 
\end{align*}
This meets all the requirements, if $x_1 \ne x_2 \ne x_3 \ne 0$. 
These conditions can be written as
\begin{align*}
r &\ne 0 \\
x_1 & \ne \frac{a+b\pm r}{3} \\
x_1 &\ne a+b\pm r \\
\end{align*}
So the problem is equivalent to finding all $a$, $b$ and $x_1$ such that $r$ is rational and
the 3 conditions are met. But I cannot show if such an $r$ exists, or if does,
enumerate some values of $a$, $b$, and $x_1$, let alone all of them.
Edit: corrected an algebraic  mistake.

Actually, your condition,
$(a+b - x_1)^2 - 4  (a \cdot b - x_1 (a+b - x_1))=r^2\tag{1}$
is quite easy to solve. Collecting powers of $x_1$, we get the equivalent,
$(a-b)^2+2(a+b)x_1-3x_1^2=r^2\tag{2}$
Assume $r=nx_1+(a-b)$ for some free variable $n$, and (2) reduces to,
$x_1(-2 a - 2 b + 2 a n - 2 b n + 3 x_1 + n^2 x_1) = 0\tag{3}$
You can then rationally solve for $x_1$ in terms of free variables $a,b,n$ (though choose them to avoid the 3 conditions you mentioned).
P.S. I forgot to add that (3) is the complete rational solution to (2). For any given $a,b,x_1$ with $x_1\neq0$ such that (2) is a square, one can always find rational n as $n = (r-(a-b))/x_1$. 
