<Question>
A steel sphere with a 3-inch radius is made by removing metal from the corners of a cube that has the shortest possible side lengths. How many cubic inches are in the volume of the cube?
</Question>

<think>
<Plan_1>
Okay, let's try to figure out this problem. So, we have a steel sphere with a 3-inch radius, and it's made by removing metal from the corners of a cube with the shortest possible side lengths. We need to find the volume of the cube. Hmm, let's break this down step by step.

<prompt_1.1> Identify the key elements of the problem and what needs to be calculated. </prompt_1.1>
</Plan_1>

<Execution_1>
<execution_1.1>
First, the sphere has a radius of 3 inches, which means its diameter is 6 inches. Since the sphere is made by removing metal from the corners of the cube, I'm guessing the sphere is inscribed inside the cube. If the sphere is inscribed, that would mean the diameter of the sphere is equal to the side length of the cube. Wait, but if the sphere is inscribed, it touches all the faces of the cube exactly once, right? So the diameter of the sphere should be equal to the edge length of the cube. Therefore, if the radius is 3, the diameter is 6, so the cube's side length would be 6 inches. Then the volume of the cube would be side length cubed, so 6^3 = 216 cubic inches.
</execution_1.1>
</Execution_1>

<Plan_2>
Based on execution_1.1:
But wait, the problem says "the shortest possible side lengths". Hmm, maybe I need to consider something else here.

<prompt_2.1> Re-evaluate the initial assumption about the sphere being inscribed in the cube. </prompt_2.1>

But then, how does the cube relate to the sphere? If the sphere is made by removing metal from the cube's corners, perhaps the sphere is the remaining part after those corners have been cut off. So maybe the sphere is inscribed in the cube, but the cube is larger, and the sphere touches the centers of the cube's faces.

<prompt_2.2> Consider the relationship between the cube and the sphere if the cube is larger and the sphere is inscribed within it. </prompt_2.2>

But wait, the problem says "the shortest possible side lengths". So maybe the cube is the smallest cube that can contain the sphere after the corners have been removed. But I'm not sure.

Alternatively, perhaps the sphere is formed by the cube with its corners cut such that the sphere is the inner remaining part. In that case, the sphere would be tangent to the original cube's faces. Wait, but if you remove metal from the corners of a cube, the remaining part is a smaller cube with its edges chamfered or something.

<prompt_2.3> Explore the concept of the cube being the smallest possible container for the sphere after corner removal. </prompt_2.3>

Wait, maybe I need to think of it as the sphere being the dual of the cube. If you start with a cube and truncate its corners (chamfer them), the resulting shape can approach a sphere, but in this problem, they are making a sphere by removing the corners. So perhaps the original cube is such that when you cut off its corners, the remaining inner cube (the part that's left after removing the corners) is the sphere. But that doesn't quite make sense because a sphere is a smooth surface, while removing corners would leave flat faces.

<prompt_2.4> Re-evaluate the problem statement and consider if the sphere is inscribed in the original cube. </prompt_2.4>

But maybe the cube is the minimal cube such that the sphere fits inside it after the corners are removed. Wait, but if the sphere is inscribed in the cube, then the cube's edges are equal to the sphere's diameter. But then, if you remove the corners of the cube, the remaining part would be a smaller cube inside, but how does that relate to the sphere?

<prompt_2.5> Consider the relationship between the cube and the sphere if the cube is minimized while containing the sphere. </prompt_2.5>
</Plan_2>

<Execution_2>
<execution_2.1>
Hold on, maybe the sphere is the part that's left after the corners are removed, so the cube must be larger than the sphere. Let me visualize this. If you take a cube and cut off each of its eight corners, the remaining part is a smaller cube in the center with its edges rounded. But in this problem, the remaining part is a sphere. So perhaps the cube is such that when you cut off its corners, the inner remaining part is a sphere with radius 3 inches.
</execution_2.1>

<execution_2.2>
Let me visualize this. If you start with a cube and cut off the corners, you end up with a smaller cube in the center where the sphere would fit. But the problem states that the sphere is made by removing metal from the cube's corners, so maybe the sphere is the remaining part after those corners are removed. In that case, the sphere would fit inside the original cube, touching the centers of each face. So the diameter of the sphere would equal the edge length of the cube. Therefore, if the radius is 3, diameter is 6, so cube side is 6. Then volume is 6*6*6=216.
</execution_2.2>

<execution_2.3>
But that might complicate things. Maybe the sphere is inscribed in the cube, meaning the cube's edges are equal to the sphere's diameter. But if that's the case, then the cube's side length is 6 inches, and the volume is 6^3 = 216 cubic inches. But then why mention removing metal from the corners? Maybe I'm misunderstanding the problem.
</execution_2.3>

<execution_2.4>
Wait, maybe the sphere is the dual? No, perhaps the sphere is the intersection of the cube with the corners truncated. Wait, perhaps the sphere is the largest possible sphere that can fit inside the cube after the corners are cut. In that case, the sphere would touch the centers of the original cube's faces. But then, the diameter of the sphere would be equal to the edge length of the cube. Wait, but if the sphere is inscribed in the cube, then the cube's edge length is equal to the sphere's diameter.
</execution_2.4>

<execution_2.5>
Wait, maybe the sphere is the dual of the cube. If you start with a cube and truncate it (cut off its corners), you can get a shape with more faces. But in this case, the sphere is created by removing metal from the cube's corners. So perhaps the original cube is larger, and when you remove the corners, the remaining part is the sphere. But how exactly?
</execution_2.5>
</Execution_2>

<Plan_3>
Based on execution_2.1, execution_2.2, execution_2.3, execution_2.4, execution_2.5:
Wait, maybe the sphere is the part that was removed from the cube. But the problem says "the steel sphere... is made by removing metal from the corners of a cube". So the sphere is the removed part, and the remaining part is the cube with its corners truncated. But then, the volume of the sphere would be equal to the volume removed from the cube. But the question is asking for the volume of the cube, not the sphere. So perhaps we need to relate the sphere's volume to the cube's volume.

<prompt_3.1> Reassess the problem statement and identify the key elements and what needs to be calculated. </prompt_3.1>
</Plan_3>

<Execution_3>
<execution_3.1>
Wait, the problem doesn't mention the volume of the sphere, but just gives its radius. Wait, the sphere is made by removing metal from the cube, so the volume of the sphere is equal to the total volume removed from the cube. But if we can find the relationship between the cube's dimensions and the sphere's dimensions, maybe we can find the cube's volume.
</execution_3.1>
</Execution_3>

<Plan_4>
Based on execution_3.1:
But the problem states the sphere has a radius of 3 inches. So the sphere's volume is (4/3)πr³ = (4/3)π(3)^3 = (4/3)π*27 = 36π cubic inches. But how does that relate to the cube's volume? If the sphere is made by removing metal from the cube's corners, the cube must have been larger than the sphere. But how much larger?

Wait, but maybe the cube is the smallest cube such that when you remove its corners, the remaining part is a sphere. But that might not make sense because removing corners from a cube would create a shape with flat faces and rounded edges, not a perfect sphere. So perhaps the problem is assuming that the sphere is perfectly inscribed in the cube, meaning the sphere touches all the faces of the cube.

<prompt_4.1> Calculate the volume of the sphere and determine if it can be inscribed in the cube. </prompt_4.1>
</Plan_4>

<Execution_4>
<execution_4.1>
In that case, the diameter of the sphere would equal the edge length of the cube. So the cube's edge length is 6 inches, volume 6*6*6=216. But then, if the sphere is perfectly inscribed, there's no metal removed from the cube; the sphere is just perfectly fitting inside the cube. So perhaps the problem is saying that the sphere is formed by removing the corners of the cube, and the cube is the smallest possible cube that allows the sphere to be formed that way. But I'm confused.
</execution_4.1>
</Execution_4>

<Plan_5>
Based on execution_4.1:
Wait, maybe "the shortest possible side lengths" refers to the minimal cube that can contain the sphere after the corners have been removed. But how does removing the corners affect the cube's size? If you remove the corners of a cube, you're essentially cutting off each vertex, which would create new faces where the corners were. The remaining part would be a convex polyhedron with both the original faces (now truncated) and new triangular faces where the corners were removed. But if the sphere is formed by removing the corners, then the sphere would fit inside the original cube, but how does that relate to the cube's dimensions?

<prompt_5.1> Consider the minimal cube that can contain the sphere after corner removal and its implications. </prompt_5.1>

Alternatively, maybe the sphere is the dual of the cube with its corners truncated. Wait, perhaps the problem is referring to the sphere being the result of rounding the cube's corners, but in this case, the sphere is the remaining part.

<prompt_5.2> Explore the concept of rounding cube corners to form a sphere and its implications. </prompt_5.2>

But how does that relate to the cube's original dimensions? If you start with a cube and chamfer its corners (cut off the corners), the remaining part is a sphere only if you chamfer them in a very specific way. But I think in reality, chamfering the corners would create a shape with flat faces and rounded edges, not a perfect sphere. So maybe the problem is using an idealized scenario where removing the corners of the cube results in a perfect sphere. In that case, the original cube must have had its corners cut such that the distance from the original cube's corner to the point where the corner is removed is such that the resulting shape is a sphere.

<prompt_5.3> Analyze the problem from the perspective of chamfering cube corners to form a sphere. </prompt_5.3>
</Plan_5>

<Execution_5>
<execution_5.1>
Wait, perhaps the cube is the smallest cube such that the sphere is entirely inside it, and the sphere touches the centers of the cube's edges or something. But I'm not sure. Maybe we need to model the cube and the sphere's relationship mathematically.

Let's consider the cube with side length 'a'. When you remove the corners of the cube, you are effectively truncating each of the eight vertices. Each corner is a right angle, and truncating it would replace the corner with a small triangular face. The amount of truncation would determine how much of the original cube is removed. The remaining part would be a polyhedron with the original six faces (each truncated) and eight new triangular faces where the corners were removed.

If the resulting shape is a sphere, that implies that the truncation is done in such a way that the resulting surface is a sphere. However, a sphere is a perfectly smooth surface, while truncating a cube's corners would create a polyhedron with flat faces. Therefore, the only way the remaining part after removing the corners could be a sphere is if the truncation is done infinitesimally, which isn't practical. Therefore, perhaps the problem is using an idealized or approximate scenario.

Alternatively, maybe the sphere is the largest possible sphere that can fit inside the cube after the corners have been removed. In that case, the sphere's size would be constrained by the cube's dimensions and the truncation. But the problem states that the sphere has a radius of 3 inches, so maybe we can use that to find the cube's side length.

Wait, perhaps the sphere is inscribed in the cube such that the cube's edges are equal to the sphere's diameter. If that's the case, then the cube's edge length is 6 inches, and the volume is 6^3 = 216. But then, why mention removing metal from the corners? Maybe the cube is the original cube before any corners are removed, and after removing the corners, the remaining part is the sphere. But that doesn't quite make sense because removing corners would create flat faces, not a sphere.

Alternatively, maybe the cube is the minimal cube that can contain the sphere after the corners have been removed. The sphere is inside the cube, and the cube's edges are aligned with the original cube's edges. The shortest possible side lengths would mean that the cube is just large enough to contain the sphere. But if the sphere is inscribed in the cube, then the cube's edge length is equal to the sphere's diameter, which is 6 inches. Therefore, the volume would be 6*6*6=216. But again, the mention of removing corners is confusing me.

Wait, perhaps the problem is not about truncating the cube's corners but rather that the sphere is formed by cutting off the cube's corners, meaning that the sphere is the part that was removed. But the problem says "the steel sphere... is made by removing metal from the corners of a cube". So the sphere is the removed part, and the remaining part is the cube with its corners cut off. In that case, the volume of the sphere would be equal to the total volume removed from the cube. But we need to find the cube's volume. However, we don't know the cube's original dimensions. So perhaps we need to relate the sphere's volume to the cube's volume through the truncation process.

But how?

Let me think. If you truncate a cube's corners, the volume removed is the volume of the eight corner tetrahedrons that are cut off. Each corner of a cube is a right angle, and truncating it involves cutting off a small octahedron or tetrahedron from each corner. Wait, if you chamfer the corner, you might be removing a small octahedron, but if you just cut off the corner with a single plane, you remove a small tetrahedron.
</execution_5.1>

<execution_5.2>
Wait, maybe the original cube is larger, and by removing the corners, you end up with a sphere. But that doesn't quite make sense. If you remove the corners of a cube, you're taking material away, but the remaining part would still have the original edges and vertices replaced by new faces. To get a sphere, you would need to remove material until the remaining part is perfectly round, which would require an infinite number of infinitesimal cuts, not just removing the corners. So perhaps the problem is using "removing the corners" in a metaphorical sense, meaning that the sphere is inscribed within the cube, touching the centers of the cube's faces. In that case, the sphere's diameter equals the cube's edge length. So if the sphere has a radius of 3, diameter 6, cube edge length 6, volume 216.
</execution_5.2>

<execution_5.3>
Wait, but a sphere is a smooth surface, while cutting off the corners of a cube would create flat faces. So unless you're removing all the corners in a way that the remaining part is a sphere, which would require an infinite number of cuts, which isn't practical. Therefore, maybe the problem is using a different interpretation. Perhaps the sphere is inscribed in the cube, and the cube is the smallest cube that can contain the sphere after the corners have been removed. But I'm not sure.

Alternatively, maybe the cube is such that the sphere is the largest possible sphere that can fit inside the cube after the corners have been removed. In that case, the sphere would be tangent to the centers of the original cube's faces. So the diameter of the sphere would be equal to the edge length of the cube. Therefore, the cube's edge length is 6 inches, volume 216. But then, why mention removing the corners? Because if the sphere is inscribed, the corners haven't been removed. So perhaps the cube's corners are removed such that the sphere is the remaining part. But how?

Wait, maybe the cube is the original cube, and when you remove the corners (chamfer them), the remaining part is a sphere. But in reality, chamfering would create flat faces, but if you chamfer all the corners of a cube in such a way that the resulting shape is a sphere, then the original cube must have had its corners cut at a very specific angle and distance. However, mathematically, this is only possible if the cube is infinite, which isn't the case here. So perhaps the problem is using a different approach.

Alternatively, perhaps the problem is referring to the cube being the dual of the sphere. Wait, but a cube and a sphere are different shapes. The dual of a cube is an octahedron. So that might not be relevant here.
</execution_5.3>
</Execution_5>

<Plan_6>
Based on execution_5.1, execution_5.2, execution_5.3:
Alternatively, perhaps the sphere is the result of rounding the cube's corners, and the cube's original dimensions are such that the rounding results in a sphere with radius 3. But how does that relate to the cube's side length?

<prompt_6.1> Reconsider the problem from the perspective of inscribing the sphere in the cube. </prompt_6.1>
</Plan_6>

<Execution_6>
<execution_6.1>
Maybe the maximum distance from the center of the cube to any corner is the original cube's space diagonal divided by 2. If the sphere is formed by rounding all the cube's corners, the maximum distance from the center to the new faces (from the rounded corners) would be equal to the sphere's radius. Wait, the original cube's space diagonal is a√3, where a is the edge length. The distance from the center to a corner is (a√3)/2. If we round the corners such that this distance becomes equal to the sphere's radius, then (a√3)/2 = 3. Solving for a gives a = (3*2)/√3 = 6/√3 = 2√3. Then, the volume of the cube would be (2√3)^3 = 8*(3√3) = 24√3. But that seems unlikely because the problem mentions the shortest possible side lengths, which might refer to the cube being minimized, but this approach might not be correct.

Alternatively, if the sphere is inscribed in the cube, touching the centers of the cube's faces, then the sphere's diameter is equal to the cube's edge length. So, diameter is 6 inches, edge length 6 inches, volume 216. But again, this doesn't involve removing the corners. So perhaps the problem is trying to say that the cube is the smallest cube that can contain the sphere, and the sphere is formed by removing the corners. But I'm still confused.

Wait, maybe the problem is that the sphere is the dual of the cube with its corners truncated. If you take a cube and truncate all its corners, the resulting shape is a truncated cube, which is an Archimedean solid. However, the problem states that the sphere is made by removing the corners, so perhaps the sphere is the limit as the truncation amount approaches the original cube's edges. In other words, if you truncate the cube's corners until the original faces are reduced to points, the remaining part would approach a sphere. But in reality, you can't get a perfect sphere by truncating a cube's corners; you can only approximate it. Therefore, maybe the problem is assuming that the sphere is a perfect fit inside the cube, touching the centers of the original cube's faces, making the cube's edge length equal to the sphere's diameter. Thus, the cube's volume is 6^3 = 216.
</execution_6.1>
</Execution_6>

<Plan_7>
Based on execution_6.1:
But then, why mention removing the corners? Unless the cube is the original cube before truncation, and the sphere is the remaining part after truncating its corners. But in that case, the sphere's volume would be the original cube's volume minus the volume of the truncated parts. However, without knowing how much of the cube is removed, we can't calculate the sphere's volume. But the problem gives the sphere's radius, so maybe we can relate the sphere's volume to the cube's volume through the truncation.

<prompt_7.1> Calculate the volume of the sphere and determine the cube's volume if the sphere is inscribed. </prompt_7.1>
</Plan_7>

<Execution_7>
<execution_7.1>
Wait, but the sphere's volume is (4/3)πr³ = 36π ≈ 113.097 cubic inches. If the sphere is inscribed in the cube, the cube's volume is 216, which is larger. But how does removing the corners relate to this? If the cube is the original cube, and you remove the corners to form a sphere, the volume removed would be the original cube's volume minus the sphere's volume. But unless we know the original cube's size, we can't compute this. Wait, but maybe the original cube's size is such that after removing the corners, the remaining part is the sphere. But how?

Alternatively, perhaps the problem is simpler. If the sphere has a radius of 3 inches, its diameter is 6 inches. If the sphere is inscribed in the cube, the cube's edges are 6 inches. Therefore, the volume is 6*6*6=216. The mention of removing corners might be a red herring, or perhaps it's meant to mislead. Maybe the problem is just asking for the cube that can inscribe the sphere, ignoring the corner removal. But that seems unlikely.

Alternatively, perhaps the cube is the smallest cube that can contain the sphere, so the cube's edges are equal to the sphere's diameter. Therefore, the answer is 216. But why mention removing metal from the cube's corners? Maybe the cube is the original cube, and after removing the corners to shape it into a sphere, the remaining part is the sphere. But unless we know how much is removed, we can't relate the two volumes. Therefore, perhaps the problem is simply asking for the cube that can inscribe the sphere, and the answer is 216. But I need to verify.

Wait, let's think again. If the sphere is made by removing metal from the cube's corners, the sphere is the remaining part. So the original cube must have been larger than the sphere. But the sphere has a radius of 3, so diameter 6. If the cube's corners are removed to form the sphere, the cube's original dimensions must be such that the sphere fits inside. But how?

Alternatively, perhaps the cube is the dual of the sphere. Wait, no. The dual of a cube is an octahedron. The sphere is a separate shape. Maybe the problem is using the term "removing metal from the corners" to mean cutting the cube to create a sphere inside. But in reality, cutting the corners of a cube doesn't create a sphere. Therefore, perhaps the problem is using a different approach, such as the cube being the smallest cube that can contain the sphere, and the sphere is inscribed. Therefore, the cube's edge length is equal to the sphere's diameter, which is 6, volume 216.
</execution_7.1>
</Execution_7>

<Plan_8>
Based on execution_7.1:
But the problem says "the shortest possible side lengths", so maybe there's a different interpretation. Perhaps the cube is the minimal cube such that the sphere fits inside it after the corners have been removed. But how?

<prompt_8.1> Re-evaluate the problem from the perspective of finding the minimal cube that can contain the sphere. </prompt_8.1>
</Plan_8>

<Execution_8>
<execution_8.1>
If you start with a cube and remove its corners, the remaining part is a smaller cube with rounded edges. The minimal cube that can contain the sphere after corner removal would have an edge length equal to the sphere's diameter. Wait, but if the sphere is inside the original cube after corner removal, the original cube's edge length must be equal to the sphere's diameter. Therefore, the cube's volume is 6^3=216. So maybe the answer is 216 cubic inches.
</execution_8.1>
</Execution_8>

<Plan_9>
Based on execution_8.1:
But I need to confirm. Let's think again.

If the sphere is made by removing metal from the corners of a cube, then the cube must have been larger than the sphere. If the sphere has a radius of 3, its diameter is 6. If the cube's corners are removed such that the remaining part is the sphere, then the cube's original edge length must be equal to the sphere's diameter. Otherwise, if the cube is smaller, the sphere would not fit. Therefore, the cube's edge length is 6, volume 216. Therefore, the answer is 216 cubic inches.

But I need to check if there's another interpretation. For example, if the cube is the original cube, and after removing the corners, the remaining part is a smaller cube inside the sphere. But that would mean the sphere is larger than the original cube, which contradicts the sphere's radius of 3. Therefore, that can't be.

Alternatively, perhaps the cube is the smallest cube that can contain the sphere, so the cube's edge length is equal to the sphere's diameter. Therefore, the volume is 6^3=216. The answer is 216.
</Plan_9>

<Execution_9>
No prompt to execute.
</Execution_9>

<Final_answer>
Therefore, the volume of the cube is $\boxed{216}$ cubic inches.
</Final_answer>
</think>