sum of all positive integral values of $a\;,$ for which equation $\lfloor x \rfloor ^3+x-a=0$ has solution 
The sum of all positive integral values of $a\;,$ Where $a\in \left[1,1500\right]$
for which the equation $\lfloor x \rfloor ^3+x-a=0$ has solution, Where $\lfloor x \rfloor $ Represent floor of $x$

$\bf{My\; Try::}$ Given $\lfloor x \rfloor ^3+x-a=0\Rightarrow x=\underbrace{a-\lfloor x \rfloor^3}_{\bf{integer\; quantity}}$
So Here  $x$ must be an $\bf{Integer\; quantity.}$
Now How can I solve after that, Help Required, Thanks

$$\lfloor x \rfloor ^3+x-a=0$$
$a\in \mathbb Z \Rightarrow x \in Z$
$$x^3+x=a$$
$$x(x^2+1)=a$$
$a\in[1;1500]$ 
Then $2\le x\le11$
If $x=2$ then $a=10$
If $x=3$ then $a=3\cdot10=30$
...
If $x=11$ then $a=11\cdot(11^2+1)=1342$
