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Question Number 119408 by ZiYangLee last updated on 24/Oct/20

If α,β,γ are the three roots of the 3x^3 +12x^2 −77x+11=0, find the value of (α−1)(β−1)(γ−1).

Ifα,β,γarethethreerootsofthe3x3+12x2−77x+11=0,findthevalueof(α−1)(β−1)(γ−1).

Commented by liberty last updated on 24/Oct/20

let p(x)=3x^3 +12x^2 −77x+11 and p(x)=0 has the roots are α, β and γ let p(x+1)=3(x+1)^3 +12(x+1)^2 −77(x+1)+11 has the roots are { ((α−1)),((β−1)),((γ−1)) :} p(x+1)=3(x^3 +3x^2 +3x+1)+12(x^2 +2x+1)−77x−77+11 p(X)=3x^3 +21x^2 −44x−51 thus (α−1)(β−1)(γ−1)=((51)/3) = 17

letp(x)=3x3+12x2−77x+11andp(x)=0hastherootsareα,βandγletp(x+1)=3(x+1)3+12(x+1)2−77(x+1)+11hastherootsare{α−1β−1γ−1p(x+1)=3(x3+3x2+3x+1)+12(x2+2x+1)−77x−77+11p(X)=3x3+21x2−44x−51thus(α−1)(β−1)(γ−1)=513=17

Answered by mr W last updated on 24/Oct/20

α+β+γ=−((12)/3) αβ+βγ+γα=−((77)/3) αβγ=−((11)/3) (α−1)(β−1)(γ−1) =αβγ+(α+β+γ)−(αβ+βγ+γα)−1 =−((11)/3)−((12)/3)+((77)/3)−1 =17

α+β+γ=−123αβ+βγ+γα=−773αβγ=−113(α−1)(β−1)(γ−1)=αβγ+(α+β+γ)−(αβ+βγ+γα)−1=−113−123+773−1=17

Commented by ZiYangLee last updated on 24/Oct/20

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