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`        If a, b, c are the roots of x3-2x2+6x-1=0, then find the value of ∑(a3+a2+a)/(a2-a+1).`
one year ago

Arun
24740 Points
```							Dear student if we find f’(x)then we know that it is always increasing fn hence it has only one real root
```
one year 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+c3 = -25.therefore the sum is {3-(-25)}/4 = 7.
```
one year ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions