Column I Column IA)If α, β are the roots of ax2 + bx + c = 0 then α3+β3α−3+β-3=P)3abc−b3c3B)If α, β are the roots of ax2 +bx + c = 0 then (aα+b)−3+(aβ+b)−3=Q)c3a3C)If α, β are the roots of ax2 +bx + c = 0 then 1α3+1β3=R)b3−3abca3c3D)If one root of ax2 + bx +c = 0 is double theother, then b2/ac = S)9/2 ABCD1)QRSP2)QRPS3)RQSP4)PQRS

Given α+β=−ba,αβ=caA) α3+β31α3+1β3=(αβ)3=ca3 B) aα2+bα+c=0α(aα+b)=−c aα+b=−cα(aα+b)−3=α3−c3 (aβ+b)−3=β3−c3(aα+b)−3+(aβ+b)−3=α3+β3−c3=(α+β)3−3αβ(α+β)−c3=−b3a3−3ca−ba−c3b3−3abca3c3 C) 1α3+1β3=α3+β3(αβ)3=(α+β)3−3αβ(α+β)(αβ)3=−b3a3−3ca−bac3a3=−b3+3abcc3=3abc−b3c3D) Let α, 2α are rootsα+2α=−ba⇒α=−b3aα(2α)=ca∴2α2=ca⇒2b29a2=ca∴b2ac=92

Column I Column IA)If α, β are the roots of ax2 + bx + c = 0 then α3+β3α−3+β-3=P)3abc−b3c3B)If α, β are the roots of ax2 +bx + c = 0 then (aα+b)−3+(aβ+b)−3=Q)c3a3C)If α, β are the roots of ax2 +bx + c = 0 then 1α3+1β3=R)b3−3abca3c3D)If one root of ax2 + bx +c = 0 is double theother, then b2/ac = S)9/2 ABCD1)QRSP2)QRPS3)RQSP4)PQRS

Given α+β=−ba,αβ=caA) α3+β31α3+1β3=(αβ)3=ca3 B) aα2+bα+c=0α(aα+b)=−c aα+b=−cα(aα+b)−3=α3−c3 (aβ+b)−3=β3−c3(aα+b)−3+(aβ+b)−3=α3+β3−c3=(α+β)3−3αβ(α+β)−c3=−b3a3−3ca−ba−c3b3−3abca3c3 C) 1α3+1β3=α3+β3(αβ)3=(α+β)3−3αβ(α+β)(αβ)3=−b3a3−3ca−bac3a3=−b3+3abcc3=3abc−b3c3D) Let α, 2α are rootsα+2α=−ba⇒α=−b3aα(2α)=ca∴2α2=ca⇒2b29a2=ca∴b2ac=92

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Column I Column I
A) If $α, β$ are the roots of ax² + bx + c = 0 then $\frac{α^{3} + β^{3}}{α^{- 3} + β^{- 3}} =$ P) $\frac{3 abc - b^{3}}{c^{3}}$
B) If $α, β$ are the roots of ax² +bx + c = 0 then $(aα + b)^{- 3} + (aβ + b)^{- 3} =$ Q) $\frac{c^{3}}{a^{3}}$
C) If $α, β$ are the roots of ax² +bx + c = 0 then $\frac{1}{α^{3}} + \frac{1}{β^{3}} =$ R) $\frac{b^{3} - 3 abc}{a^{3} c^{3}}$
D) If one root of ax² + bx +c = 0 is double the
other, then b²/ac = S) 9/2
A B C D
1) Q R S P
2) Q R P S
3) R Q S P
4) P Q R S

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answer is B.

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Detailed Solution

Given $α + β = \frac{- b}{a}, αβ = \frac{c}{a}$
$A) \frac{α^{3} + β^{3}}{\frac{1}{α^{3}} + \frac{1}{β^{3}}} = (αβ)^{3} = {(\frac{c}{a})}^{3} B) {aα}^{2} + bα + c = 0 α (aα + b) = - c aα + b = \frac{- c}{α} (aα + b)^{- 3} = \frac{α^{3}}{- c^{3}} \begin{aligned} (aβ + b)^{- 3} = \frac{β^{3}}{- c^{3}} (aα + b)^{- 3} + (aβ + b)^{- 3} = \frac{α^{3} + β^{3}}{- c^{3}} \\ = \frac{(α + β)^{3} - 3 αβ (α + β)}{- c^{3}} = \frac{\frac{- b^{3}}{a^{3}} - 3 \frac{c}{a} (\frac{- b}{a})}{- c^{3}} \frac{b^{3} - 3 abc}{a^{3} c^{3}} \end{aligned} \begin{aligned} C) \frac{1}{α^{3}} + \frac{1}{β^{3}} = \frac{α^{3} + β^{3}}{(αβ)^{3}} = \frac{(α + β)^{3} - 3 αβ (α + β)}{(αβ)^{3}} \\ = \frac{\frac{- b^{3}}{a^{3}} - \frac{3 c}{a} (\frac{- b}{a})}{\frac{c^{3}}{a^{3}}} = \frac{- b^{3} + 3 abc}{c^{3}} = \frac{3 abc - b^{3}}{c^{3}} \end{aligned}$
D) Let $α$ , 2 $α$ are roots
$\begin{array}{l} α + 2 α = \frac{- b}{a} \Rightarrow α = \frac{- b}{3 a} \\ α (2 α) = \frac{c}{a} ∴ 2 α^{2} = \frac{c}{a} \Rightarrow 2 (\frac{b^{2}}{9 a^{2}}) = \frac{c}{a} ∴ \frac{b^{2}}{ac} = \frac{9}{2} \end{array}$

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