1729

My solution is similar to Aareyan's, but becoming a bit more transparent through the use of substitutions.

Let $A=a+1, B=b+1,C=c+1,D=d+1$ We are told that $ABC=2, BCD=8, ACD=3, ABD=10$ Multiplying these together, we find that $(ABCD)^3=2*8*3*10=480$ .

Now $(A-B)CD=3-8=-5, (A-D)BC=2-8=-6, (C-B)AD=-7, (C-D)AB=-8$ Multiplying these together, we find that $(A-B)(A-D)(C-B)(C-D)(ABCD)^2=5*6*7*8=1680$

Finally, $((A-B)(A-D)(C-B)(C-D))^3=\frac{1680^3}{480^2}=\boxed{20580}$

add one to each eqn to get: $(a+1)(b+1)(c+1)=2$ $(b+1)(c+1)(d+1)=8$ $(c+1)(d+1)(a+1)=3$ $(d+1)(a+1)(b+1)=10$ multiply all and square both sides to get $((a+1)(b+1)(c+1)(d+1))^6=(2*8*3*10)^2$ now subtract the original equation given as pairs: $(a-d)(bc+c+b+1)=(d-a)(b+1)(c+1)=1-9=-8$ $(a-b)(cd+c+d+1)=(b-a)(c+1)(d+1)=1-7=-6$ $(c-b)(ad+a+d+1)=(b-c)(a+1)(d+1)=2-7=-5$ $(c-d)(ab+a+b+1)=(d-c)(a+1)(b+1)=2-9=-7$ multiply these and cube both sides to get $((a-b)(a-d)(c-b)(c-d))^3((a+1)(b+1)(c+1)(d+1))^6=1680^3$ substitute value : $((a-b)(a-d)(c-b)(c-d))^3=\dfrac{1680^3}{(2*8*3*10)^2}=\boxed{20580}$