the number of real values for x such that{$\displaystyle (x+1)^3$}$\displaystyle = x^3$ where {$\displaystyle x$} is the fractional part of x

i.e{2.635}=0.635

i think the answer is 1 cuz we get the answer right when we substitute 0 and i cannot think any other values that satisfy this equation

options are:

(A)0

(B)1

(C)4

(D)6