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Question

If α,β,γ are the zeroes of the cubic polynomial x3+4x+2, then find the value of
1α+β+1β+γ+1γ+α

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Solution

We have,
P(x)=x3+4x+2

Since, α,β,γ are the zeroes of this polynomial.

Now,
α+β+γ=01=0

αβ+βγ+γα=41=4

αβγ=21=2

Since,
1α+β+1β+γ+1γ+α

(β+γ)(γ+α)+(α+β)(γ+α)+(α+β)(β+γ)(α+β)(β+γ)(γ+α)

βγ+αβ+γ2+αγ+αγ+α2+βγ+αβ+αβ+αγ+β2+βγ(α+β)(β+γ)(γ+α)

α2+β2+γ2+3(αβ+βγ+γα)(α+β)(β+γ)(γ+α)

α2+β2+γ2+2(αβ+βγ+γα)+(αβ+βγ+γα)(α+β)(β+γ)(γ+α)

(α+β+γ)2+(αβ+βγ+γα)(α+β)(β+γ)(γ+α)

0+0(α+β)(β+γ)(γ+α)

0

Hence, this is the answer.

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