If α,β,γ are the roots of the equation x3 – px2 + qx + r = 0 then (α+β)(β+γ)(γ+α)=

α+β=x⇒α+β+γ−γ=x⇒p−γ=x⇒γ=p−xcollect product of roots of f(p-x) = 0⇒(p-x)3-pp-x2+qp-x+r=0⇒p3-x3-3p2x+3px2-pp2-2px+x2+pq-qx+r=0⇒x3+1-3px2+xq+3p2-pq-r=0product of roots =pq+r

If α,β,γ are the roots of the equation x3 – px2 + qx + r = 0 then (α+β)(β+γ)(γ+α)=

α+β=x⇒α+β+γ−γ=x⇒p−γ=x⇒γ=p−xcollect product of roots of f(p-x) = 0⇒(p-x)3-pp-x2+qp-x+r=0⇒p3-x3-3p2x+3px2-pp2-2px+x2+pq-qx+r=0⇒x3+1-3px2+xq+3p2-pq-r=0product of roots =pq+r

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If $α, β, γ$ are the roots of the equation x³ – px² + qx + r = 0 then $(α + β) (β + γ) (γ + α) =$

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$r - p q$

$2 (p^{2} - 3 q)$

$\frac{p^{2} - 2 q}{r^{2}}$

$q p + r$

answer is C.

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Detailed Solution

$\begin{aligned} α + β = x \Rightarrow α + β + γ - γ = x \\ \Rightarrow p - γ = x \Rightarrow γ = p - x \end{aligned}$

collect product of roots of f(p-x) = 0
$\Rightarrow {(p - x)}^{3} - p {(p - x)}^{2} + q (p - x) + r = 0$
$\Rightarrow p^{3} - x^{3} - 3 p^{2} x + 3 p x^{2} - p (p^{2} - 2 p x + x^{2}) + p q - q x + r = 0$
$\Rightarrow x^{3} + (1 - 3 p) x^{2} + x (q + 3 p^{2}) - p q - r = 0$
$p r o d u c t o f r o o t s = p q + r$