# cube of an integer

For proving the assertion, one needs only to check the identity

 $a^{3}\;\equiv\;\left(\frac{(a\!+\!1)a}{2}\right)^{\!2}-\left(\frac{(a\!-\!1)a}% {2}\right)^{\!2}.$

For example we have  $(-2)^{3}=1^{2}\!-\!3^{2}$  and  $4^{3}=64=10^{2}\!-\!6^{2}$.

Summing the first $n$ positive cubes, the identity allows http://planetmath.org/encyclopedia/TelescopingSum.htmltelescoping between consecutive brackets,

 $\displaystyle 1^{3}\!+\!2^{3}\!+\!3^{3}\!+\!4^{3}\!+\ldots+\!n^{3}$ $\displaystyle\;=\;[1^{2}\!-\!0^{2}]\!+\![3^{2}\!-\!1^{2}]\!+\![6^{2}\!-\!3^{2}% ]\!+\![10^{2}\!-\!6^{2}]\!+\ldots+\!\left[\!\left(\frac{(n\!+\!1)n}{2}\right)^% {\!2}\!-\!\left(\frac{(n\!-\!1)n}{2}\right)^{\!2}\right]$ $\displaystyle\;=\;\left(\frac{n^{2}\!+\!n}{2}\right)^{\!2},$

saving only the square $\left(\frac{(n+1)n}{2}\right)^{2}$.  Thus we have this expression presenting the sum of the first $n$ positive cubes (cf. the Nicomachus theorem  ).

Title cube of an integer CubeOfAnInteger 2013-03-22 19:34:33 2013-03-22 19:34:33 pahio (2872) pahio (2872) 11 pahio (2872) Theorem msc 11B37 msc 11A25 NicomachusTheorem TriangularNumbers DifferenceOfSquares