positive integers of the form pq, and only those of the form pq,
have 4 factors, 1,p,q, and pq  where p and q are different primes.

So let A = pq

Positive integers of the form rs², and only those of the form rs²,
have 6 factors, 1,r,s,s²,rs and rs²  where r and s are different primes.

So let B = ra²

Positive integers of the form tuv, and only those of the form tuv,
have 8 factors, 1,t,u,v,tu,tv,uv and tuv  where t, u and v are 
3 different primes.

So let D = tuv

Since D = AB/C, C = AB/D = pqrs²/(tuv)

So we have 

A = pq  we can assume p < q
B = ra²
C = pqrs²/(tuv)
D = tuv we can assume t < u < v

We will try to use only the smallest three primes, {2,3,5} if possible.
If so, then A and B can be made smaller than D which is the product
of three different primes.

We make D as small as possible by choosing t=2, u=3, v=5, so D=30.
We make B as small as possible by choosing r=3, s=2, so B=3×2² = 12.
We would like to make A as small as possible, which would be 6 with p=2 and q=3.
But that won't do, because since v=5, C = pqrs²/(tuv), then pqrs² must
be divisible by 5, and since neither r nor s are 5, the larger of p,q 
must be 5, so q=5, and we can choose p=2.  So A = pq = 2×5 = 10, and
C = pqrs²/(tuv) = 2×5×3×2²/(2×3×5) = 4

So A+B+C+D = 10+12+4+30 = 56, the smallest possible sum.

Edwin