document.write( "Question 990042: prove that n^3-n is always divided by 6 \n" ); document.write( "

Algebra.Com's Answer #610171 by htmentor(1343) $\"\"$ $\"About$

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We can factor n^3 - n as:
\n" ); document.write( "n^3 - n = n(n^2-1) = n(n-1)(n+1)
\n" ); document.write( "So for every number n, we have the product of three consecutive integers.
\n" ); document.write( "For example:
\n" ); document.write( "3,4,5
\n" ); document.write( "6,7,8
\n" ); document.write( "12,13,14 etc.
\n" ); document.write( "In all these cases, there will be at least one even number, which is divisble by 2.
\n" ); document.write( "And in any trio of consecutive numbers, there will always be one number which is divisible by 3.
\n" ); document.write( "Thus the product of any three consecutive integers has factors of 2 and 3, which means it is divisible by 6.
\n" ); document.write( " \n" ); document.write( "

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