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Question

Let p, q, r be roots of cubic equation x3+2x2+3x+3=0, then

A
pp+1+qq+1+rr+1=6
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B
(pp+1)3+(qq+1)3+(rr+1)3=44
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C
(pp+1)3+(qq+1)3+(rr+1)3=38
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D
pp+1+qq+1+rr+1=5
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Solution

The correct option is D pp+1+qq+1+rr+1=5
Using transformation
Let pp+1=yp=y1y
p is a root of the given cubic equation.
So p3+2p2+3p+3=0
(y1y)3+2(y1y)2+3(y1y)+3=0
y35y2+6y3=0
So y1,y2,y3 are root of above equation, so
pp+1+qq+1+rr+1=5
yi=5,yiyj=6,yiyjyk=3
y31+y32+y333y1y2y3=(yi)[(yi)23yiyj]
y31+y32+y333(3)=5(523(6))
(pp+1)3+(qq+1)3+(rr+1)3=44

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