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Question

# If α,β,γ are the roots of the cubic x3−px2+qx−r=0, find the equations whose roots are (β+γ−α),(γ+α−β),(α+β−γ)

A
y3ry2+(4rq2)y+p(8r4pq+p3)=0
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B
y3py2+(4qp2)y+p(8r4pq+p3)=0
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C
ry3q(q+1)y2+p(r+1)2y(p+1)3=0
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D
ry3q(r+1)y2+p(r+1)2y(r+1)3=0
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Solution

## The correct option is B y3−py2+(4q−p2)y+p(8r−4pq+p3)=0As α,β,γ are roots of x3−px2+qx−r=0Then S1=α+β+γ=p,S2=αβ+βγ+γα=+q,S3=αβγ=rNow S′1=∑(α+β−γ)=∑(p−2γ)∴(β+γ−α)+(γ+α−β)+(α+β−γ)=pS′3=∏(α+β−γ)=(p−2γ)(p−2α)(p−2β)=(p−2γ)(p2−(2α+2β)p+4αβ)=p3−(2α+2β)p2+4αβp−2γp2+(2α+2β)2γp−8αβγ=p3−(2α+2β+2γ)p2+p(4αβ+4αγ+4βγ)−8αβγ=p3−2p3+4pq−8r=−8r+4pq−p3And S′2=(α+β−γ)(γ+α−β)+(β+γ−α)(γ+α−β)+(β+γ−α)(α+β−γ)=(p−2γ)(p−2α)+(p−2α)(p−2β)+(p−2β)(p−2γ)=p2−(2α+2γ)p+4αγ+p2−(2α+2β)β+4αβ+p2−(2α+2γ)p+4γβ=(4q−p2)Therefore using y3+(−S′1)y2+(S′2)y+(−S3)=0 required equation isy3−py2+(4q−p2)y+p(8r−4pq+p3)=0

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