The volume of the solid obtained by rotating the region enclosed by
$$y=1/x^3,y=0,x=2,x=4$$
about the line $x=-4$ can be computed using the method of cylindrical shells via an integral
$$V={\int }_a^b◻?$$
with limits of integration $a=$ and $b=$

Step 1 ：The volume of a solid obtained by rotating a region about a line can be computed using the method of cylindrical shells. The formula for the volume is given by: $$V=2\pi {\int }_a^{b}r(x)h(x)dx$$ where $r(x)$ is the radius of the cylindrical shell at $x$ and $h(x)$ is the height of the cylindrical shell at $x$.

Step 2 ：In this case, the region is bounded by $y=1/x^3$, $y=0$, $x=2$, and $x=4$, and is rotated about the line $x=-4$.

Step 3 ：The radius of the cylindrical shell at $x$ is the distance from $x$ to the line $x=-4$, which is $x-(-4)=x+4$.

Step 4 ：The height of the cylindrical shell at $x$ is the distance from $y=0$ to $y=1/x^3$, which is $1/x^3-0=1/x^3$.

Step 5 ：So, we have $r(x)=x+4$ and $h(x)=1/x^3$.

Step 6 ：The limits of integration are the values of $x$ that bound the region, which are $x=2$ and $x=4$.

Step 7 ：So, the integral that gives the volume of the solid is: $$V=2\pi {\int }_2^4(x+4)(1/x^3)dx$$

Step 8 ：Now, we can compute this integral to find the volume. The final answer is $\boxed{{\displaystyle \frac{5\pi }{4}}}$.