Proof without words: The difference of consecutive cubes

Algebra II, Theorems 1 Minute

Source: https://www.facebook.com/photo.php?fbid=451328078334029&set=a.416585131808324.1073741827.416199381846899&type=1&theater

For a more conventional algebraic proof, notice that

(n+1)^3 - n^3 = n^3 + 3n^2 + 3n + 1 - n^3 = 3n(n+1) + 1

The product n(n+1) is always an even number times an odd number: if n is even, then n+1 is odd, but if n is odd, then n(n+1). So n(n+1) is a multiple of 2, and so 3n(n+1) is a multiple of 6. Therefore, 3n(n+1)+1 is one more than a multiple of 6, proving the theorem.

Unknown's avatar

Published by John Quintanilla

I'm a Professor of Mathematics and a University Distinguished Teaching Professor at the University of North Texas. For eight years, I was co-director of Teach North Texas, UNT's program for preparing secondary teachers of mathematics and science.

Published

Leave a comment

This site uses Akismet to reduce spam. Learn how your comment data is processed.