The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is

We need to find the volume of a sphere that contains a rectangular box inscribed inside it.

When a rectangular box is inscribed in a sphere, all eight corners of the box touch the sphere's surface. This means the longest diagonal of the box (called the body diagonal) passes through the center and equals the sphere's diameter.

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Let the dimensions of the rectangular box be $l$, $b$, and $h$.

From the given information:

Surface area of box = $2(lb + bh + hl) = 846$ sq cm

Sum of all 12 edges = $4(l + b + h) = 144$ cm

A rectangular box has 4 edges of length $l$, 4 edges of length $b$, and 4 edges of length $h$.

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From surface area: $lb + bh + hl = 423$

From edge sum: $l + b + h = 36$

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We need $l^2 + b^2 + h^2$ to find the body diagonal $\sqrt{l^2 + b^2 + h^2}$.

Using the identity: $(l + b + h)^2 = l^2 + b^2 + h^2 + 2(lb + bh + hl)$

When we expand $(l + b + h)^2$, we get all the square terms plus all the cross products doubled.

Therefore: $l^2 + b^2 + h^2 = (l + b + h)^2 - 2(lb + bh + hl)$

$l^2 + b^2 + h^2 = (36)^2 - 2(423)$

$= 1296 - 846$

$= 450$

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Body diagonal = $\sqrt{l^2 + b^2 + h^2} = \sqrt{450}$

$450 = 9 \times 50$

$= 9 \times 25 \times 2$

$= 225 \times 2$

$\sqrt{450} = \sqrt{225 \times 2} = 15\sqrt{2}$

Diameter of sphere = $15\sqrt{2}$

Radius of sphere = $\frac{15\sqrt{2}}{2}$

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Volume of sphere = $\frac{4}{3}\pi R^3$

Volume = $\frac{4}{3}\pi \left(\frac{15\sqrt{2}}{2}\right)^3$

$\left(\frac{15\sqrt{2}}{2}\right)^3 = \frac{(15\sqrt{2})^3}{2^3}$

$= \frac{15^3 \times (\sqrt{2})^3}{8}$

$= \frac{3375 \times 2\sqrt{2}}{8}$

$= \frac{6750\sqrt{2}}{8}$

Volume = $\frac{4}{3}\pi \times \frac{6750\sqrt{2}}{8}$

$= \frac{4\pi \times 6750\sqrt{2}}{24}$

$= \frac{27000\pi\sqrt{2}}{24}$

$= 1125\pi\sqrt{2}$