Drawing a picture is always a good thing to do for these types of problems.
Start by sketching the graphs of $y=8$, $y=x^3$, and $x=0$. We then see that the region to be revolved about the $x$-axis is in the first quadrant. It is bounded above by the line $y=8$, on the left by the $y$-axis, and on the right by the graph of $y=x^3$. Note that the point of intersection in the upper right is $(2,8)$ and can be found by solving the equation $8=x^3$.
The region is shown, shaded in blue, below:

Now, you generate the solid by revolving this region about the $x$-axis. The cylindrical shells are generated by revolving a horizontal line segment, shown in green in the diagram, at a fixed $\color{maroon}y$-value about the $x$-axis.
These line segments "start" at $y=0$ and "end" at $y=8$. Thus the integral giving the volume of the solid of revolution is with respect to $y$ and is of the form $\int_{y=0}^{y=8} 2\pi \,\color{maroon}{ r_y} \cdot\color{darkgreen}{ h_y} \, dy$ where $\color{maroon}{r_y}$ is the radius of the shell at $\color{maroon}y$ and $\color{darkgreen}{h_y}$ is the height of the shell at $\color{maroon}y$.
The height of the shell at $\color{maroon}y$ is the length of the line segment at $\color{maroon}y$. Since the length of the line segment at $\color{maroon}y$ is the $x$-coordinate of its right hand endpoint, we have
$\ \ \ \ \ \ \ \color{darkgreen}{h_y}=\color{darkgreen}{y^{1/3}}$.
Keep in mind that we want to write $\color{darkgreen}{h_y}$ in terms of $y$, since the integral is with respect to $y$.
The radius of the shell is the height above the $x$-axis of the line segment:
$\ \ \ \ \ \ \ \color{maroon}{r_y}=\color{maroon}y$.
Thus, the volume of the solid of revolution is: $ \int_0^8 2\pi\,\color{maroon}y\cdot\color{darkgreen}{ y^{1/3}} \, dy =\int_0^8 2\pi\cdot y^{4/3} \, dy= \textstyle{6\pi\over7} y^{7/3}\Bigr|_0^8={6\pi\over7}\cdot 8^{7/3} ={6\pi\over7} \cdot2^{7 }= {6\pi\over7} \cdot128={768\pi\over 7}. $