If the roots of the equation x^3-ax^2+bx-c=0 are three consecutive integers, then what is the smallest possible value of b ?

• Smallest possible value of b = -1.
  
  
  Let the roots be x-1,x,x+1
  
  b = x(x-1) + x(x+1) + x^2 -1
  = x^2 - x + x^2 + x + x^2 -1
  = 3x^2 - 1
  
  Now f(x) = b = 3x^2 -1
  
  f'(x) = 6x = 0 => x =0
  
  f''(x) = 6 => x = 0 is minima
  
  Hence, min b = -1
• 12 years agoHelpfull: Yes(3) No(1)