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If x1, x2, x3, xn are the roots of x^n+a...

If `x_1, x_2, x_3, x_n` are the roots of `x^n+ax+b=0` then the value of `(x_1-x_2)(x_1 -x_3) (x_1-x_4)...... (x_1-x_n)` is equal to

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`P(x)=(x-x_1)(x-x_2)(x-x_3)...(x-x_n)=x^n+ax+b`
`lim_(x->x_1)(x-x_2)(x-x_3)...(x-x_n)=lim_(x->x_1)((x^n+ax+b)/(x-x_1))`
`(x-x_2)(x-x_3)...(x_1-x_n)=lim_(x->x_1)(x^n+ax+b)/(x-x_1)`
`lim_(x->x_1)(nx^(n-1)+a)/1`
`nx_1^(n-1)+a`
option B is correct.
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