1
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Question

# If α,β & γ are the roots of the equation x3−x−1=0 then the value of, 1+α1−α+1+β1−β+1+γ1−γ is

A
Zero
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B
1
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C
7
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D
1
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Solution

## The correct option is D 1We have, Cubic equation is x3−x−1=0 Now, we can written as, x3+0x2−x−1=0 Roots are α,βandγ Then, Sum of roots =−coeff.ofx2coeff.ofx3 α+β+γ=−01 α+β+γ=0......(1) Both product of roots=coeff.ofxcoeff.ofx3 αβ+βγ+γα=−11 αβ+βγ+γα=−1......(2) Product of roots=−constanttermcoeff.ofx3 αβγ=−11 αβγ=−1......(3) From equation (1), (2) and (3) to and we get, α=1,β=−1,γ=0 Then, According to question, 1+α1−α+1+β1−β+1+γ1−γ ⇒1+11−1+1−11+1+1+01−0 ⇒1 Hence, this is he answer.

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