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Question

If α,β,γ are the roots of x3+2x23x+1=0, then

A
αβα+β+αγα+γ+βγβ+γ=117
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B
αβα+β+αγα+γ+βγβ+γ=117
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C
1α+1β+1γ=3
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D
1α+1β+1γ=3
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Solution

The correct option is D 1α+1β+1γ=3
Given equation is x3+2x23x+1=0 has roots as α,β,γ
Now, assuming
S=αβα+β+αγα+γ+βγβ+γS=11α+1β+11α+1γ+11β+1γ
Finding the equation whose roots are 1α,1β,1γ
x33x2+2x+1=0
Here, 1α+1β+1γ=3,
So, S=131γ+131β+131α
S=⎢ ⎢ ⎢ ⎢11γ3+11β3+11α3⎥ ⎥ ⎥ ⎥

Finding the equation whose roots are 1α3,1β3,1γ3
(x+3)23(x+3)2+2(x+3)+1=0x3+6x2+11x+7=0

Finding the equation whose roots are 11α3,11β3,11γ3
7x3+11x2+6x+1=0

Therefore, S=(117)
S=117

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