How do i figure out how many combination's of crosses if there is a trihybrid cross?

trihybrid crosses can get somewhat complicated. lets say you're crossing AaHHJj  x  aahhJJ

what you need to do is take the first letter on each set of letters and go one by one through the genes until you've gone through each.

for example,
AaHHJj  x aahhJJ

for AaHHJj, you would have AHJ, AHj, aHJ, and aHj
for aahhJJ, you would just have ahJ

what you have to really do is take take the first letter add another letter onto it depending on the next allele. so AaHHJj, you take the A, then add an H, then add a J, making it AHJ. then the next one you do AH and this time a lower case j because it is different from the first one somehow.

you always want to make sure you have one of each and never repeat the same sequences twice

If I understand the question correctly, the number of possible gametes  is given by the formula 2?(number of heterozygous genes), so that, since AaBbCc has 3 heterozygous genes, 2?3 = 8.  Therefore, AaBbCc produces 8 possible gametes.  A trihybrid cross of AaBbCc x AaBbCc would produce an 8x8 Punnett square or 64 offspring.
A genotype of AAbbcc produces 1 kind of gamete, since there are 0 heterozygous genes, and 2?0 = 1;  that gamete is Abc.  The cross AAbbCCDDee x AABbCCddee looks complex, but it's only a 1x2 Punnett square...