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Mathematical melody strange but true…..

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  • Kurien Thomas
    A. (a+b+c+d+e)^2 = a^+b^+c^3+d^3+e^3 The above equation at a glance looks incorrect. B. Can a cube of any number be expressed as the difference of two squares?
    Message 1 of 1 , Jul 14, 2005
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      A. (a+b+c+d+e)^2 = a^+b^+c^3+d^3+e^3

      The above equation at a glance looks incorrect.

      B. Can a cube of any number be expressed as the difference of two squares?

      A is possible when a, b, c, d, e are consecutive natural numbers starting from 1 ie 1, 2, 3, 4, 5. B is possible because of this property of A.
      (1+2+3+4+5)^2=1^3+2^3+4^3+5^3
      15^2=1+8+27+64+125
      ie 225=225
      From the above it is evident that
      (1+2+3+4+5+…+n)^2 =1^3+2^3+3^3+4^3+5^3+…+n^3

      10^3= 55^2-45^2
      This can be derived from the above formula. Consider
      (1+2+3+4+…+10)^2=1^3+2^3+3^3+4^3+…+10^3…..x
      (1+2+3+4+…+9)^2=1^3+2^3+3^3+4^3+…+9^3……..y

      Subtracting y from x
      x-y=55^2-45^2=10^3

      so 36^3 can be expressed as
      36^3= (1+2+3+4+…+36)^2-(1+2+3+4+…+35)^2
      = 36^3=666^2-630^2=(666+630)x(666-630)=(1296x36)=(36^2x36)=36^3

      So a cube of any number having 10 digits, 20 digits…. so on can be expressed in the above form…

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