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+y⇒p=y1−y
As p is a root of the given equation, so
p3+2p2+3p+3=0⇒(y1−y)3+2(y1−y)2+3(y1−y)+3=0⇒y3+2y2(1−y)+3y(1−y)2+3(1−y)3=0⇒y3−5y2+6y−3=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)2−3∑y1y2]=3⋅3+(5)(52−3⋅6)=44