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Both relators have zero exponent sums, so abelianization gives $H_1(Y)={\mathbb{Z}}^3$, with meridian variables $x,y,z$ corresponding to $a,b,c$. The universal abelian cover has deck transformation group ${\mathbb{Z}}^3$ and coefficient group ringThe generalized Heegaard diagram gives a two-dimensional spine with one vertex, three edges and two faces. Its lifted cellular chain complex iswhere chosen lifts of the cells giveThe two columns are the abelianized Fox derivatives of $[a,bc]$ and $[ab,c]$. The Fox calculus identity gives $d_1{d}_2=0$; this can also be checked by multiplying the displayed matrices. There is no three-cell in this spine. One may use the lifted spine because its deformation retraction from $Y$ lifts to the universal abelian cover.

The maximal minors of the Alexander matrix, in row-pair order $(1,2),(1,3),(2,3)$, areTheir greatest common divisor in $R$ is $xyz-1$, since $x-1,y-1,z-1$ have no common nonunit divisor. Accordingly the multivariable Alexander polynomial isThe allowed units are $\pm x^i{y}^j{z}^k$. The single-variable specialization convention can introduce extra $(t-1)$ factors; the answer here is the genuinely multivariable Alexander polynomial.

Keep all four relators and all five generators, in the orders $(r_A,r_B,r_C,r_D)$ and $(d,e,a,b,c)$. Under abelianization, these generators become $(t,t,t^2,t^3,t^2)$. The Fox calculus rules areFor example, $r_B=a{d}^{-1}c{b}^{-1}$ gives the nonzero Fox derivative entries $-a{d}^{-1}$, $1$, $-a{d}^{-1}c{b}^{-1}$, $a{d}^{-1}$ in columns $d,a,b,c$, respectively. After abelianization these are $-t,1,-1,t$. The full Alexander matrix isThe Fox calculus identity supplies a useful check:Deleting the $e$ column, adjacent to the unbounded region across the marked meridian of a knot, givesThus $\det B$ is a representative of the Alexander polynomial of a knot of the figure-eight knot, with its customary ambiguity of multiplication by $\pm t^n$.