document.write( "Question 242460: Two positive integers differ by 6 and their squares differ by 72. By how much do
\n" ); document.write( "their cubes differ?
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Algebra.Com's Answer #177479 by JimboP1977(311)\"\" \"About 
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\"a-b+=+6\"
\n" ); document.write( "\"a%5E2-b%5E2+=+72\"
\n" ); document.write( "\"a%5E3-b%5E3+=+x\"\r
\n" ); document.write( "\n" ); document.write( "Well the way I did it which may not be the best, but here goes:\r
\n" ); document.write( "\n" ); document.write( "a^2-a^b can also be written as (a+b)(a-b). We know that a-b = 6\r
\n" ); document.write( "\n" ); document.write( "So (a+b)*6 = 72 so a+b = 72/6 = 12 , 2a-2b = 12.\r
\n" ); document.write( "\n" ); document.write( "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.\r
\n" ); document.write( "\n" ); document.write( "The first two numbers that came to mind were 9 and 3. And these are indeed correct.\r
\n" ); document.write( "\n" ); document.write( "So 9^3 - 3^3 = 702
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