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Question

# If α,β,γ are the roots of the equation x3+px2+qx+r=0,r≠0 and βγ+1α, αγ+1β, αβ+1γ are the roots of the equation x3+ax2+bx+c=0,c≠0, then

A
a=q(1r)r
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B
b=p(1r)2r
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C
c=(1r)3r
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D
ab=pq(1r)
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Solution

## The correct option is C c=(1−r)3r x3+px2+qx+r=0 ...(1) x3+ax2+bx+c=0 ...(2) Let m be a root of eqn(1) and n be a root of eqn(2). n=βγ+1α=αβγ+1α=1−rα [∵αβγ=−r] ⇒n=1−rm⇒m=1−rn Substitute m=1−rn in eqn(1), we get (1−rn)3+p(1−rn)2+q(1−rn)+r=0 ⇒rn3+q(1−r)n2+p(1−r)2n+(1−r)3=0 Replacing n by x and dividing by r (∵r≠0), we get x3+q(1−r)rx2+p(1−r)2rx+(1−r)3r=0 Therefore, a=q(1−r)r, b=p(1−r)2r, c=(1−r)3r

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