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I really want to explain the reasoning behind the calculations, so I apologize if this gets rather wordy! The beauty of the mathematics involved is that you only need to understand the probabilities for a single trait cross in order to determine the probabilities for multiple trait crosses. Because the traits are assumed to assort independently here (mathematically, they are independent events), the overall probability is the product of the individual probabilities.
In a monohybrid cross, the probability of the two heterozygous parents having a heterozygous offspring is 1/2. This probability is derived as follows, using alleles A and a as examples:
Dad has a 1/2 chance of passing on the A allele, and a 1/2 chance of passing on the a allele. These are the only two alleles possible, and he can only give one at a time. The same can be said for mom: 1/2 chance of A, and 1/2 chance of a. The assortment of alleles in mom and dad are independent events (dad's assortment doesn't affect mom's assortment and vice versa), so the probability of alleles assorting and coming together is the product of the independent probabilities: 1/2 x 1/2 = 1/4. However, this takes into account only one type of heterozygote, with dad contributing A and mom contributing a, for example. A heterozygote is also produced by dad contributing a and mom contributing A, which again has a 1/4 probability. Because we don't care which parent actually contributed which allele, the probabilities can be added together, so that the probability of either aA or Aa is 1/4 + 1/4 = 2/4 = 1/2.
So now that we know the reasoning behind the 1/2 chance of two heterozygotes having a heterozygous child, this is all we need to concentrate on at this point. If the probability of heterozygosity is 1/2 for a single cross between two heterozygotes, the probability of heterozygosity between parents who are triply heterozygous for traits of interest is the product of the single trait crosses. In this case, that means 1/2 x 1/2 x 1/2 = 1/8. Because all the alleles are assumed to assort independently (that is, the assortment of A and a has no effect on the assortment of B and b or C and C), each cross is an independent event, and the probabilities can be multiplied.
So after all of that, the probability of two parents heterozygous for three unlinked traits having a child that is also heterozygous is 1/8.
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