Vedic Sutra #8:     By the Completion or Non-Completion

We want to solve for x from:
  x3 + Ax2 + Bx + C = 0

Example #1:
    x3 + 6x2 - 37x + 30 = 0

  1. Recognize x3 + 6x2 as the first two terms of (x+2)3, so we rewrite the equation as:
        x3 + 6x2 = 37x - 30
  2. "Complete the Cube" and (rearranging terms) we get the equation:
        (x+2)3 = [12x + 8] + 37x - 30   ... adding those extra terms [12x + 8] to the right-side   Note: (x+2)3 = x3 + 6x2 + 12x + 8
  3. Now let x + 2 = y (or x = y - 2) and get (after simplifying):
        y3 - 49y + 120 = 0
    ... and now the quadratic term, y2, is missing.
  4. If this has a root y = c, then it must be factorable, like so:
        (y - c)(y2 + cy - 120/c)   ... so that the y2 term cancels and the constant term is 120
  5. We note that, for c = 3, we'd get:
        (y - 3)(y2 + 3y - 40)   ... which gives the correct constant term when multiplied out
  6. Factoring, again, we get:
        (y - 3)(y - 5)(y + 8) = 0   so   y = 3, 5 or 8
  7. Since x = y - 2, then   x = 1, 3 or 6.

Sometimes, "Completing the Cube" doesn't use the first two terms to suggest the appropriate completion ...

Example #2:
    x3 + 7x2+ 14x + 8 = 0

  1. Recognize x3 + 9x2 as the first two terms of (x+3)3 = x3 + 9x2 + 27x + 27.
    We'll use this completion and rewrite the equation as:
        (x+3)3 = [27x + 27] + 2x2 - 14x + 8   ... adding the extra terms [27x + 27] to the right-side
  2. Then let x + 3 = y (or x = y - 3) and get ... eventually!  
        (y - 1)(y + 1)(y - 2) = 0
  3. So
        y = 1, -1 or 2   so x = -2, -4 or -1.
Why?
The method (as I see it) relies heavily upon recognizing certain combinations of terms, pairs of related numbers and, most particularly, a certain genetic disposition

... or divine intervention