Question: Let $M$ and $N$ be $3$-manifolds obtained by zero-surgery on (left-handed) trefoil and figure-eight knot respectively.

What is the easy way to prove that $M$ and $N$ are not homeomorphic?

Note: When they are knot homology spheres (they are both homology $S^1 \times S^2$'s), I cannot use the classical invariants.

Answer: 
If you're happy bringing in heavy machinery then you could compute some sort of Floer homology, like the 'hat' version of Heegaard Floer homology: this has rank 2 for $S^3\_0(3\_1)$ and rank 4 for $S^3\_0(4\_1)$, so they're different.

On the other hand, in cases like this where you have a very specific pair of 3-manifolds in mind, it's often enough to distinguish their fundamental groups by counting covers of some finite order. Here you could work this out by hand starting from Wirtinger presentations of the respective knot groups, or you could just ask SnapPy to do it:

```
In[1]: len(Manifold('3_1(0,1)').covers(9))
Out[1]: 4

In[2]: len(Manifold('4_1(0,1)').covers(9))
Out[2]: 2

```

This counts 9-fold covers of each 0-surgery, and since the numbers are different, they must have different fundamental groups.
As long as we're collectively throwing the kitchen sink at this, note that the Alexander polynomial of the knot is an invariant of the 0-surgered manifold. So since the figure eight and trefoil knots have different polynomials, the 0-surgeries are not homeomorphic.