Find the volume of the solid rotating about $x=2$ $y=x^3, y=0, x=1$ rotated about $x=2$
Plugging in $x=1$ into $y=x^3$ we get 1 so our bounds are $[0,1]$
In order to account for rotating about the line $x=2$ subtract 2 from the function
$$\int_0^1 \pi [y^{\frac{1}{3}}-2]^2 dy$$
$$\pi \int_0^1 y^{\frac{2}{3}}-4y^{\frac{1}{3}}+4$$
$$\frac{3}{5}y^{\frac{5}{3}}-3y^{\frac{4}{3}}+4y \Big\vert_0^1$$
pluggin all of this in I get:
$\frac{3}{5}+1=\frac{8}{5} \pi$
The answer is supposed to be $\frac{3\pi}{5}$

For any given $y$-value, what is the area of the corresponding annulus ("washer") cross section of our solid of rotation? The outer radius is $2 - y^{1/3}$, while the inner radius is $1$. So the area becomes
$$
\pi(2-y^{1/3})^2 - \pi\cdot 1^2 = \pi\left[(2-y^{1/3})^2 - 1\right]
$$
This is what you should integrate. All in all, it's not difficult to see that the additional $-1$ decreases your answer by $\pi$, which means we get the result we should.
Your calculation didn't take this inner radius into account, and thus calculated the volume of the region between $y= x^3$, $y = 0$, $y = 1$ and $x = 2$, rotated about $x = 2$. This has the extra square bounded by $y = 0, y = 1, x = 1, x = 2$ that shouldn't have been included. Note that the solid of rotation of this region is a cylinder with volume $\pi$.
