When I've got a proof like this I find it helpful to consider an example first.
Let's say a = 48 = 2 x 2 x 2 x 2 x 3
and b = 56 = 2 x 2 x 2 x 7
The hcf = 2 x 2 x 2
and lcm = 2 x 2 x 2 x 2 x 3 x 7
so hcf x lcm = (2 x 2 x 2) x (2 x 2 x 2 x 2 x 3 x 7) = (2 x 2 x 2 x 2 x 3) x (2 x 2 x 2 x 7) = 48 x 56
The common factors occur in both the hcf and lcm and the not-common factors occur in the lcm. So the re-arrangement allows us to pick out one set of common factors together with one set of not-common factors for the first number and what is left is the factors for the other. Here's an attempt to make that rigorous:
Suppose a = hcf x N where N is the not-common factors, and similarly b = hcf x M
Then the lcm = all the common factors once and the not-common factors from both = hcf x N x M
So hcf x lcm = (hcf) x ( hcf x N x M) = (hcf x N) x ( hcf x M) = a x b
Hope that helps,
Bob
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