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If `alpha_1, ,alpha_2, ,alpha_n` are the roots of equation `x^n+n a x-b=0,` show that `(alpha_(1)-alpha_(2))(alpha_(1)-alpha_(3))...(alpha_(1)-alpha_(n))=nalpha_(1)^(n-1)+nalpha`

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Since `alpha_(1),alpha_(2),…,alpha_(n)` are the roots of equation
`x^(n)+nax-b=0`, we have
`x^(n)+nax-b=(x-alpha_(1))(x-alpha_(2))…(x-alpha_(n))`
`(x^(n)+nax-b)/(x-alpha_(1))=(x-alpha_(2))(x-alpha_(3))...(x-alpha_(n))`
or `underset(xtoalpha_(1))lim(x^(n)+nax-b)/(x-alpha_(1))=underset(xtoalpha_(1))lim[(x-alpha_(2))(x-alpha_(3))...(x-alpha_(n))]`
or `underset(xtoalpha_(1))lim(nx^(n-1)+na)/(1)=(alpha_(1)-alpha_(2))(alpha_(1)-alpha_(3))...(alpha_(1)-alpha_(n))`
(Using L'Hospital's rule on LHS)
or `(alpha_(1)-alpha_(2))(alpha_(1)-alpha_(3))...(alpha_(1)-alpha_(n))=nalpha_(1)^(n-1)+nalpha`
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