#### Answer

Yes, one-to-one.

#### Work Step by Step

We are given:
$f(x)=3+x^{3}$
This function is one-to-one because it passes the horizontal line test (it is a shifted cubic function, which is odd). We can show this algebraically.
We start with the assumption:
$x_1\neq x_2$
$x_{1}^{3}\neq x_{2}^{3}$
$3+x_{1}^{3}\neq 3+x_{2}^{3}$
Because cubing unique values produces another unique value. Thus the function will not have the same $y$ value for two different $x$ values. It is therefore one-to-one.