If $α, β, γ$ are different from 1 and are the roots of $a x^{3} + b x^{2} + c x + d = 0$ and $(β - γ) (γ - α) (α - β) = 25 / 2$ , then the determinant $Δ = | \begin{matrix} \frac{α}{1 - α} & \frac{β}{1 - β} & \frac{γ}{1 - γ} \\ α & β & γ \\ α^{2} & β^{2} & γ^{2} \end{matrix} |$ equals

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$\frac{- 25 d}{2 (a + b + c + d)}$

$\frac{25 d}{a}$

$\frac{25 d}{2 a}$

$\frac{- 25 d}{a + b + c + d}$

answer is D.

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Detailed Solution

Taking $α, β, γ$ common from $C_{1}, C_{2}, C_{3}$ respectively, we get

$Δ = α β γ | \begin{matrix} \frac{1}{1 - α} & \frac{1}{1 - β} & \frac{1}{1 - γ} \\ 1 & 1 & 1 \\ α & β & γ \end{matrix} |$ $= α β γ | \begin{matrix} \frac{1}{1 - α} & \frac{1}{1 - β} - \frac{1}{1 - α} & \frac{1}{1 - γ} - \frac{1}{1 - α} \\ 1 & 0 & 1 \\ α & β - α & γ - α \end{matrix} |$
[using $C_{2} \to C_{2} - C_{1}$ , and $C_{3} \to C_{3} - C_{1}$ ]
$= \frac{α β γ (- 1) (β - α) (γ - α)}{(1 - α) (1 - β) (1 - γ)} | \begin{matrix} 1 - γ & 1 - β \\ 1 & 1 \end{matrix} |$

$= \frac{α β γ (α - β) (β - γ) (γ - α)}{(1 - α) (1 - β) (1 - γ)}$

As $α, β, γ$ are the roots of $a x^{3} + b x^{2} + c x + d = 0$ .
$a x^{3} + b x^{2} + c x + d = a (x - α) (x - β) (x - γ)$
And $α β γ = - d / a$
Thus, $Δ = \frac{(- d / a) (25 / 2)}{(a + b + c + d) / a} = - \frac{25 d}{2 (a + b + c + d)}$