How to show that $\pi_1(\mathbb{R}^3\setminus k_{2,3}\#-k_{2,3})\not\cong\pi_1(\mathbb{R}^3\setminus k_{2,3}\# k_{2,3})$? How to show that $\pi_1(\mathbb{R}^3\setminus k_{2,3}\#-k_{2,3})\not\cong\pi_1(\mathbb{R}^3\setminus k_{2,3}\# k_{2,3})$ by their presentations? 
Here $k_{2,3}$ stands for the trefoil knot. $k_{2,3}\#-k_{2,3}$ stands for the connected sum of $k_{2,3}$ with its mirror image, and $k_{2,3}\# k_{2,3}$ stands for the connected sum of $k_{2,3}$ with itself.
I have no difficulty finding their presentations:
$$\pi_1(\mathbb{R}^3\setminus k_{2,3}\#-k_{2,3})\cong\langle a_1,a_2,a_3,a_4\mid a_1a_2a_1=a_2a_1a_2,\ a_3a_4a_3=a_4a_3a_4,\ a_2=a_4a_3a_4^{-1}\rangle$$
$$\pi_1(\mathbb{R}^3\setminus k_{2,3}\#k_{2,3})\cong\langle b_1,b_2,b_3,b_4\mid b_1b_2b_1=b_2b_1b_2,\ b_3b_4b_3=b_4b_3b_4,\ b_2=b_4^{-1}b_3b_4\rangle$$
But I'm in great trouble showing that they are not isomorphic. I rarely dealt with such problems, and the long complex equations are disturbing to me. By the way, they are really not isomorphic, right?
Any hint, method or solution is welcomed. 
 A: It may come as a surprise: these two groups are isomorphic!
Here's the general idea (with a concrete calculation with your presentations at the end).  For a connect sum $K_1\mathbin{\#} K_2$ of knots, there is a ball $B\subset S^3$ whose boundary $\partial B$ intersects $K_1\mathbin{\#} K_2$ in exactly two points, where $B\cap (K_1\mathbin\# K_2)$ is the "$K_1$" part of the connect sum.  The complement $X_1=B- \nu(K_1\mathbin\# K_2)$ is homeomorphic to $S^3-\nu(K_1)$, preserving orientations, where $\nu(K_1\mathbin\# K_2)\subset S^3$ denotes a tubular neighborhood of the connect sum.  Similarly, $X_2=\overline{S^3-B}-\nu(K_1\mathbin\# K_2)$ is homeomorphic, preserving orientations, to $S^3-\nu(K_2)$.  ($\overline{S^3-B}$ is homeomorphic to a closed ball.)
So, $S^3-\nu(K_1\mathbin\# K_2)=X_1\cup X_2$ with $X_1\cap X_2$ being an annulus $A$. By the van Kampen theorem, $\pi_1(S^3-\nu(K_1\mathbin\# K_2))$ is the amalgamated free product $\pi_1(X_1)*_A\pi_1(X_2)$.  The image of the generator for $\pi_1(A)$ in $\pi_1(X_1)$ and $\pi_1(X_2)$ are two consistently-oriented meridians for the respective knot complement.  (An "oriented meridian" is one that links positively with the knot, given the knot's orientation.)
Let $-\overline{K_2}$ indicate the mirror image of the inverse of the knot.  The inverse is the knot with inverted orientation (remember: the connect sum of two knots depends on the orientation of the two knots in general --- many small knots, like the trefoil, are isotopic to their inverse).  The mirror image map $S^3\to S^3$ induces an isomorphism $\pi_1(S^3-K_2)\to \pi_1(S^3-(-\overline{K_2}))$ that sends an oriented meridian to an oriented meridian.
With $\mu_1$ and $\mu_2$ being respective meridians for $K_1$ and $K_2$, $$\pi_1(S^3-\nu(K_1\mathbin\#K_2))\approx\left(\pi_1(S^3-K_1)*\pi_1(S^3-K_2)\right)/\langle \mu_1\mu_2^{-1}\rangle$$
Using the above comments about mirror images of inverses, this is also
$$\left(\pi_1(S^3-K_1)*\pi_1(S^3-(-\overline{K_2}))\right)/\langle \mu_1\mu_2^{-1}\rangle\approx \pi_1(S^3-\nu(K_1\mathbin\# -\overline{K_2}))$$
In your case, the trefoil $k_{2,3}$ is isotopic to its inverse,
so
$$\pi_1(S^3-k_{2,3}\mathbin\#k_{2,3}) \approx \pi_1(S^3-(k_{2,3}\mathbin\#-k_{2,3})).$$
(I'm just using your notation.  I think mirror images are usually denoted with an overline or by prefixing with an $m$.)
Concretely, there is an isomorphism
$$\pi_1(S^3-(k_{2,3}\mathbin\#-k_{2,3})) \to \pi_1(S^3-k_{2,3}\mathbin\#k_{2,3}) $$ which we can give in the following way.
Since I don't knot exactly how you got your particular presentations, I recomputed the fundamental groups:

Notice that $a_1a_3=a_3a_2=a_2a_4=a_4a_3$, so $a_1=a_4$.  This corresponds to the generator of the fundamental group of the annulus separating the two knot exteriors:

Using the first three relations, we can calculate $a_1a_2a_1=a_2a_1a_2$, and with the last three, $a_1a_5a_1=a_5a_1a_5$.  Hence,
$$\pi_1(S^3-(k_{2,3}\mathbin\#k_{2,3}))\approx \langle a_1,a_2,a_5\mid a_1a_2a_1=a_2a_1a_2,a_1a_5a_1=a_5a_1a_5\rangle.$$
Doing the same thing for the second knot, you get
$$\pi_1(S^3-(k_{2,3}\mathbin\#-k_{2,3}))\approx \langle b_1,b_2,b_5\mid b_1b_2b_1=b_2b_1b_2,b_1b_5b_1=b_5b_1b_5\rangle.$$
So the homomorphism defined by $a_1\mapsto b_1$, $a_2\mapsto b_2$, and $a_3\mapsto b_3$ is an isomorphism.

Waldhausen proved that $\pi_1(S^3-K)$, along with the peripheral subgroup generated by a meridian and a longitude, determines $K$.  Your example shows why the fundamental group of the complement is not sufficient.
