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Question

Let p,q,r be the roots of x3+2x2+3x+3=0, then which of following is/are correct?

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

The correct option is B (pp+1)3+(qq+1)3+(rr+1)3=44
x3+2x2+3x+3=0 has roots as p,q,r
Let pp+1=y, so
p=py+yp=y1y

As p is a root of the given equation, so
p3+2p2+3p+3=0(y1y)3+2(y1y)2+3(y1y)+3=0y3+2y2(1y)+3y(1y)2+3(1y)3=0y35y2+6y3=0
Whose roots are y1=pp+1,y2=qq+1,y3=rr+1

Now,
Sum of roots is
pp+1+qq+1+rr+1=5

(pp+1)3+(qq+1)3+(rr+1)3=(y1)3+(y2)3+(y3)3=3y1y2y3+(y1)[(y1)23y1y2]=33+(5)(5236)=44

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