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Question

If α,β,γ are the roots of the cubic x32x+3=0, then value of 1α3+β3+6+1β3+γ3+6+1γ3+α3+6 is equal to

A
13
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B
13
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C
12
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D
12
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Solution

The correct option is B 13
Since, α,β,γ are the roots of the equation
α32α+3=0
β32β+3=0
γ32γ+3=0
α3+β3=2(α+β)6
β3+γ3=2(β+γ)6
γ3+α3=2(γ+α)6
1α3+β3+6+1β3+γ3+6+1γ3+α3+6=12(α+β)+12(β+γ)+12(γ+α)
=12(γ)+12(α)+12(β)=αβ+βγ+αγ2αβγ
=22(3)=13

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