Using properties of determinants, find the following:
$∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ α^{2} & β^{2} & γ^{2} β + γ & γ + α & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$

(α + β) (β + γ) (γ - α) (α + β + γ)

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(α - β) (β - γ) (γ - α) (α + β + γ)

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None of these

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Solution

The correct option is B $(α - β) (β - γ) (γ - α) (α + β + γ)$
$∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ α^{2} & β^{2} & γ^{2} β + γ & γ + α & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$
$C_{1} \to C_{1} - C_{2}$
$C_{2} \to C_{2} - C_{3}$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} α - β & β - γ & γ α^{2} - β^{2} & β^{2} - γ^{2} & γ^{2} β - α & γ - β & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$
$= ∣ ∣ ∣ ∣ ∣ \begin{matrix} α - β & β - γ & γ (α - β) (α + β) & (β - γ) (β + γ) & γ^{2} - (α - β) & - (β - γ) & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$
Taking out common $(α - β), (β - α)$ from $C_{1}, C_{2}$ respectively.
$= (α - β) (β - γ) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 1 & 1 & γ (α + β) & (β + γ) & γ^{2} - 1 & - 1 & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$
$R_{1} \to R_{1} + R_{3}$
$= (α - β) (β - γ) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & α + β + γ (α + β) & (β + γ) & γ^{2} - 1 & - 1 & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$
Taking out common $(α + β + γ)$ from $R_{1}$
$= (α - β) (β - γ) (α + β + γ) ∣ ∣ ∣ ∣ ∣ \begin{matrix} 0 & 0 & 1 (α + β) & (β + γ) & γ^{2} - 1 & - 1 & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣$
Expanding along $R_{1}$
$= (α - β) (β - γ) (γ - α) (α + β + γ)$

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Similar questions

∣ ∣ ∣ ∣ ∣ \begin{matrix} α & α^{2} & β + γ β & β^{2} & γ + α γ & γ^{2} & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣

= (α - β) (β - γ) (γ - α) (α + β + γ)

Using properties of determinants, prove that:

∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ α^{2} & β^{2} & γ^{2} β + γ & γ + α & α + β \end{matrix} ∣ ∣ ∣ ∣ ∣ = x ∣ ∣ ∣ ∣ ∣ \begin{matrix} α & β & γ α^{2} & β^{2} & γ^{2} 1 & 1 & 1 \end{matrix} ∣ ∣ ∣ ∣ ∣

α, β, γ

x^{3} + 2 x^{2} + 3 x + 8 = 0

(α + β) (β + γ) (γ + α) =

α

β

γ

x^{3} - p x^{2} + q x + r = 0

(α + β) (β + γ) (γ + α) =

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