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  4. Let α, β, γ be the roots of the equation (x-a)(x-b)(x-c)=d,(d ≠ 0), then the roots of the equation (x-α)(x-β)(x-γ)+d=0, is

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Q. Let $\alpha, \beta, \gamma$ be the roots of the equation $(x-a)(x-b)(x-c)=d,(d \neq 0)$, then the roots of the equation $(x-\alpha)(x-\beta)(x-\gamma)+d=0$, is

Complex Numbers and Quadratic Equations

Solution:

$( x - a )( x - b )( x - c )- d =( x -\alpha)( x -\beta)( x -\gamma)$
$\Rightarrow( x -\alpha)( x -\beta)( x -\gamma)+ d =( x - a )( x - b )( x - c ) $
$ \Rightarrow a , b , c$