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Grade: 12
        
If a, b, c are the roots of x3-2x2+6x-1=0, then find the value of ∑(a3+a2+a)/(a2-a+1).
7 months ago

Answers : (2)

Arun
20430 Points
							
Dear student
 
if we find f’(x)
then we know that it is always increasing fn hence it has only one real root
 
 
7 months ago
saket
12 Points
							
we have, (a3+a2+a)/(a2-a+1) = a(a2+a+1)/(a2-a+1) = a(a3-1)/{(a2-a+1)(a-1)} = a(a3-1)/(a3-2a2+2a-1)
= a(a3-1)/(a3-2a2+6a-1-4a) = (1-a3)/4,               [ since ‘a’ is a root of x3-2x2+6x-1, a3-2a2+6a-1=0 ]
similarly (b3+b2+b)/(b2-b+1) = (1-b3)/4 and 
              (c3+c2+c)/(c2-c+1) = (1-c3)/4.
therefore the required sum is {3-(a3+b3+c3)}/4.
now,
       a3+b3+c3-3abc = (a+b+c){(a+b+c)2-3(ab+bc+ca)}
also a+b+c=2, ab+bc+ca=6, and abc=1.
hence a3+b3+c= -25.
therefore the sum is {3-(-25)}/4 = 7.
 
7 months ago
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