To find the volume of the sphere enclosing a rectangular box, we are given: the box's surface area is 846 sq cm, and the sum of its edge lengths is 144 cm.
Let the box's dimensions be \(a\), \(b\), and \(c\). The problem yields these equations:
- Surface Area: \[2(ab+bc+ca)=846 \rightarrow ab+bc+ca=423\]
- Edge Sum: \[4(a+b+c)=144 \rightarrow a+b+c=36\]
Since the box is inscribed in a sphere, the box's diagonal is the sphere's diameter. The diagonal \(d\) can be found using the 3D Pythagorean theorem:
\[d=\sqrt{a^2+b^2+c^2}\]
We use the algebraic identity:
\[(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\]
Substituting the known values:
\[36^2=a^2+b^2+c^2+2 \times 423\]
\[1296=a^2+b^2+c^2+846\]
Solving for \(a^2+b^2+c^2\):
\[a^2+b^2+c^2=450\]
The sphere's diameter is therefore:
\[d=\sqrt{450}=15\sqrt{2}\]
The sphere's radius \(r\) is half the diameter: \(r=\frac{d}{2}=\frac{15\sqrt{2}}{2}\).
The sphere's volume \(V\) is calculated using the formula:
\[V=\frac{4}{3}\pi r^3\]
Substituting \(r=\frac{15\sqrt{2}}{2}\):
\[V=\frac{4}{3}\pi \left(\frac{15\sqrt{2}}{2}\right)^3\]
\[=\frac{4}{3}\pi \times \left(\frac{3375 \times 2\sqrt{2}}{8}\right)\]
\[=\frac{4}{3}\pi \times \frac{6750\sqrt{2}}{8}\]
\[=\pi \times 1125\sqrt{2}\]
The volume of the sphere is \(\boxed{1125\pi\sqrt{2}}\) cubic cm.