If x1, x2, x3⋯xn are roots of xn+ax+b=0, then the value of (x1−x2)(x1−x3)(x1−x4)⋯(x1−xn) =
n(x1)n−1+a
xn+ax+b=(x−x1)(x−x2)(x−x3)⋯(x−xn)⇒ xn+ax+bx−x1=(x−x2)(x−x3)(x−x4)⋯(x−xn)Now limx→x1xn+ax+bx−x1=limx→x1(x−x2)(x−x3)(x−x4)⋯(x−x4)∴ nx1+a=(x1−x2)(x1−x3)(x1−x4)⋯(x1−xn)