# cube of an integer

Theorem. Any cube of integer is a difference of two squares, which in the case
of a positive cube are the squares of two successive triangular numbers^{}.

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,

${1}^{3}+{2}^{3}+{3}^{3}+{4}^{3}+\mathrm{\dots}+{n}^{3}$ | $\mathrm{\hspace{0.33em}}=[{1}^{2}-{0}^{2}]+[{3}^{2}-{1}^{2}]+[{6}^{2}-{3}^{2}]+[{10}^{2}-{6}^{2}]+\mathrm{\dots}+\left[{\left({\displaystyle \frac{(n+1)n}{2}}\right)}^{2}-{\left({\displaystyle \frac{(n-1)n}{2}}\right)}^{2}\right]$ | ||

$\mathrm{\hspace{0.33em}}={\left({\displaystyle \frac{{n}^{2}+n}{2}}\right)}^{2},$ |

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 |
---|---|

Canonical name | CubeOfAnInteger |

Date of creation | 2013-03-22 19:34:33 |

Last modified on | 2013-03-22 19:34:33 |

Owner | pahio (2872) |

Last modified by | pahio (2872) |

Numerical id | 11 |

Author | pahio (2872) |

Entry type | Theorem |

Classification | msc 11B37 |

Classification | msc 11A25 |

Related topic | NicomachusTheorem |

Related topic | TriangularNumbers |

Related topic | DifferenceOfSquares |