# Let g(x) be a polynomial with leading coefficient 1, whose three roots are the reciprocal of the three roots of f(x) = x`^3` + ax`^2` + bx + c, where 1< a<b.

`g(1) ` is `1 + ( a + b + 1 ) / c .`

## Expert Answers Denote the roots of `f ( x ) ` as `u , v , w . ` Because the leading coefficient of `f ` is `1 , ` we can state that `f ( x ) = ( x - u ) ( x - v ) ( x - w ) . ` Because of this, `c = -uvw , ` `b = uv + vw + uw , ` `a = - ( u + v + w ) ` (Vieta's theorem).

Further, because `g ` also has the leading coefficient `1 ` and the roots `1 / u , 1 / v , 1 / w , ` it is equal to `( x - 1 / u ) ( x - 1 / v ) ( x - 1 / w ) .`

Now consider `g ( 1 ) = ( 1 - 1 / u ) ( 1 - 1 / v ) ( 1 - 1 / w ) =`

`= 1 / ( uvw ) ( ( u - 1 ) ( v - 1 ) ( w - 1 ) ) = 1 / ( uvw ) ( uvw - ( uv + vw + uw ) + ( u + v + w ) - 1 ) ,` which is equal to `-1 / c ( -c - b - a - 1 ) = 1 + ( a + b + 1 ) / c .`

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