#### Answer

$h(x)$ is one-to-one.

#### Work Step by Step

The function $h(x)=x^3+8$ is one-to-one because it passes the horizontal line test (a shifted $x^3$ graph). We can show this algebraically:
We start with the assumption that:
$x_1\ne x_2$
Then:
${x_1}^3\ne {x_2}^3$
(Since cubing a unique number results in a unique value.)
${x_1}^3+8\ne {x_2}^3+8$
So the function would not have the same $y$ value for two different $x$ values (one-to-one).