Show that α=³√(7+5√2)+³√(7-5√2) is an integer. No calculators or computers allowed!

We start with c = cbrt[7+5sqrt2] + cbrt[7-5sqrt2].

Cube both sides:
c^3 = (7+5sqrt2) + (7-5sqrt2)
      + 3cbrt[(7+5sqrt2)^2*(7-5sqrt2)]
      + 3cbrt[(7+5sqrt2)*(7-5sqrt2)^2]
c^3 = 14 + 3*cbrt[-7-5sqrt2] + 3*cbrt[-7+5sqrt2]
c^3 = 14 + (-3)*(cbrt[7+5sqrt2] + 3*cbrt[7-5sqrt2])

Substitute c for (cbrt[7+5sqrt2] + cbrt[7-5sqrt2]) on the right side:
c^3 = 14 - 3c
c^3 + 3c - 14 = 0
(c - 2)(c^2 + 2c + 7) = 0
c = 2 is the only real solution.

Edited on October 13, 2004, 1:07 pm

Posted by Brian Smith on 2004-10-13 13:06:00