Biology help - trihybrid cross punnett squares?

You need to include every possible combination of alleles for the three genes for each parent. Fortunately, both are heterozygous for all three, so you only have to figure this out once.

The easiest way is to "do it in binary." You have three "bits," each bit being a gene. Count in three bits from 000 to 111:

000 001 010 011 100 101 110 111

Now, substitute the alleles for the digits. If '0' represents the dominant allele and '1' the recessive, and each bit represents one of the genes, you get your gametes:

HGT HGt HgT Hgt hGT hGt hgT hgt

Since both parents are the same genotype, they produce the same gametes, so there is no need to repeat for the other parent. Now, you can just build your square (which will have 64 squares):

 . . . . . HGT . . . . HGt . . . .HgT . . . . Hgt . . . hGT . . . .hGt . . . hgT . . . hgt
HGT . HHGGTT HHGGTt HHGgTT HHGgTt HhGGTT HhGGTt HhGgTT HhGgTt
HGt . .HHGGtT. HHGGtt. HHGgtT. HHGgtt HhGGtT. HhGGtt. HhGgtT. HhGgtt
HgT . .HHgGTT. HHgGTt. HHggTT. HHggTt. HhgGTT. HhgGTt HhggTT. HhggTt
Hgt . . HHgGtT . HHgGtt . HHggtT . HHggtt. HhgGtT. HhgGtt . HhggtT. Hhggtt
hGT . .hHGGTT. hHGGTt. hHGgTT hHGgTt hhGGTT hhGGTt. hhGgTT. hhGgTt
hGt . .hHGGtT . hHGGtt . hHGgtT. hHGgtt. hhGGtT. hhGGtt . hhGgtT. hhGgtt
hgT . .hHgGTT . hHgGTt . hHggTT. hHggTt. hhgGTT. hhgGTt . hhggTT. hhggTt
hgt . . hHgGtT . hHgGtt. . hHggtT. hHggtt. . hhgGtT. hhgGtt. . hhggtT . hhggtt

As you can see, with more than two genes Punnett squares quickly become cumbersome. Three genes have 8 possible gametes, four genes have 16, five have 32, etc. Usually, the forked line method is used for predicting phenotypic outcomes with more than two genes, which I'm sure you come across soon if you haven't already.