SOLUTION: Two positive integers differ by 6 and their squares differ by 72. By how much do their cubes differ?

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Question 242460: Two positive integers differ by 6 and their squares differ by 72. By how much do
their cubes differ?
Answer by JimboP1977(311) About Me  (Show Source):
You can put this solution on YOUR website!
a-b+=+6
a%5E2-b%5E2+=+72
a%5E3-b%5E3+=+x
Well the way I did it which may not be the best, but here goes:
a^2-a^b can also be written as (a+b)(a-b). We know that a-b = 6
So (a+b)*6 = 72 so a+b = 72/6 = 12 , 2a-2b = 12.
So a+b = 2a-2b, so 3b=a. From this we can gather that we are looking for two postive integers with a difference of 6, with one of them 3 times greater than the other.
The first two numbers that came to mind were 9 and 3. And these are indeed correct.
So 9^3 - 3^3 = 702