If x  is real, the function (x−a)(x−b)(x−c)  will assume all real values, provided:

Let y=(x−a)(x−b)(x−c) or      y(x−c)=x2−(a+b)x+ab or      x2−(a+b+y)x+ab+cy=0             Δ=(a+b+y)2−4(ab+cy)             =y2+2y(a+b−2c)+(a−b)2 Since x is real and y  assumes all real values, we must have Δ≥0  for all real values of y . The sign of a quadratic in y  is same as of first term provided its discriminant B2−4AC<0 This will be so it 4(a+b−2c)2−4(a−b)2<0 or     4(a+b−2c+a−b)(a+b−2c−a+b)<0 or       16(a−c)(b−c)<0  or 16(c−a)(c−b)=−υe ∴     c  lies between a  and b , i..e., a