Okay, let's try to figure out how many cylindrical beakers are needed to fill a spherical tank. Hmm, first, I need to remember the formulas for the volumes of a cylinder and a sphere. The volume of a cylinder is πr²h, right? And the volume of a sphere is (4/3)πR³. 

So the problem is asking, how many times the volume of the cylindrical beaker (which is 8 cm high and radius 3 cm) fits into the volume of the spherical tank with radius 6 cm. 

Let me start by calculating the volume of the cylinder. The radius is 3 cm, so r = 3. The height h is 8 cm. Plugging into the formula, volume_cylinder = π*(3)²*8. Let me compute that. 3 squared is 9, multiplied by 8 gives 72. So volume_cylinder = 72π cm³. 

Next, the volume of the sphere. The radius is 6 cm, so R = 6. Using the formula (4/3)πR³. Let's compute R³ first. 6 cubed is 6*6*6 = 216. Then multiply by 4/3: (4/3)*216. Let's see, 216 divided by 3 is 72, multiplied by 4 is 288. So volume_sphere = 288π cm³. 

Now, to find out how many beakers are needed, I need to divide the volume of the sphere by the volume of the cylinder. That is, number_of_beakers = volume_sphere / volume_cylinder. Plugging in the numbers: 288π / 72π. The π cancels out, so 288/72. Let me compute that. 72*4 = 288, so 288 divided by 72 is 4. 

Wait, so does that mean exactly 4 beakers? That seems straightforward. Let me just double-check my calculations to make sure I didn't make a mistake. 

Cylinder volume: π*3²*8 = π*9*8 = 72π. Sphere volume: (4/3)π*6³ = (4/3)π*216 = 288π. Dividing 288π by 72π gives 4. Yep, that checks out. 

Is there any catch here? Like, does the shape of the tank affect the number of beakers needed? The problem says "to fill a spherical tank of radius 6 cm", and the beakers are cylindrical. So the question is purely about volume, regardless of shape, since we're just transferring water from one container to another. So the number of beakers should just depend on the total volume of water divided by the volume per beaker. 

Therefore, the answer should be 4. But let me just confirm once more. 

If each beaker holds 72π cm³, and the tank holds 288π cm³, then 288π divided by 72π is indeed 4. So yes, 4 beakers. 

I think that's all. The key was to use the volume formulas correctly and then divide them. Volume problems often require this kind of straightforward division once the formulas are applied. No complicated geometry here since we're just matching volumes. 

So the final answer should be 4. Let me just write that in the box as requested.
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