Question about an integral.. Why is $a = $2? I found the item in monbukagakusho 2013 math B exam.

Consider the function
  $$F(x) = \int_a^x f(t)\ dt = x^3 - 2x^2 - x - a$$
  with $a \ne 0$. Find $a$.

I looked at the answer sheet and $a = 2$, but I ended up with weird equation (?)
$0 = a^3-2a^2-2a$
so what should I do to get $a = 2$?
Thank you 

Differentiating both sides of the equation, with respect to $x$, you get 
$$f(x) = 3x^2 - 4x - 1$$
Now you want to integrate $f$ between $a$ and $x$; this yields
$$F(x) = \int_a^x 3t^2 - 4t - 1 \ dt = |t^3 - 2t^2 - t|_a^x = x^3 - 2x^2 - x - a^3 + 2a^2 + a$$
But you want $F(x) =  x^3 - 2x^2 - x  + a$; this implies $-a^3   + 2a^2 = 0$, and given that $a \neq 0$, the only solution is $a = 2$
