Using Vieta's Formula? (3)

Algebra Level 5

2017 x 3 + 2016 x 2 + 2018 x + 2016 = 0 \large 2017x^3+2016x^2+2018x+2016=0

Let α \alpha , β \beta and γ \gamma be the roots of the equation above.

Find the value of :

2017 ( α 3 − β 3 − γ 3 ) + 2016 ( α 2 − β 2 − γ 2 ) + 2018 ( α − β − γ ) 2 ( α 6 β 7 γ 7 + α 7 β 6 γ 7 + α 7 β 7 γ 6 ) + ( α 8 β 6 γ 6 + α 6 β 8 γ 6 + α 6 β 6 γ 8 ) 8 \large \dfrac{2017(\alpha^3-\beta^3-\gamma^3)+2016(\alpha^2-\beta^2-\gamma^2)+2018(\alpha-\beta-\gamma)}{\sqrt[8]{2(\alpha^{6}\beta^{7}\gamma^{7}+\alpha^{7}\beta^{6}\gamma^{7}+\alpha^{7}\beta^{7}\gamma^{6})+(\alpha^{8}\beta^{6}\gamma^{6}+\alpha^{6}\beta^{8}\gamma^{6}+\alpha^{6}\beta^{6}\gamma^{8})} }


The answer is 2017.

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Tommy Li by Tommy Li , 22, Hong Kong

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1 solution

Tommy Li
Jul 18, 2016

{ 2017 α 3 + 2016 α 2 + 2018 α = − 2016 2017 β 3 + 2016 β 2 + 2018 β = − 2016 2017 γ 3 + 2016 γ 2 + 2018 γ = − 2016 \begin{cases} 2017\alpha^{3}+2016\alpha^{2}+2018\alpha=-2016 \\ 2017\beta^{3}+2016\beta^{2}+2018\beta=-2016 \\ 2017\gamma^{3}+2016\gamma^{2}+2018\gamma=-2016 \end{cases}

α + β + γ = − 2016 2017 \alpha+\beta+\gamma = -\frac{2016}{2017}

α β γ = ( − 1 ) 3 2016 2017 = − 2016 2017 \alpha\beta\gamma = (-1)^3\frac{2016}{2017} = -\frac{2016}{2017}

2016 ( α 3 − β 3 − γ 3 ) + 2017 ( α 2 − β 2 − γ 2 ) + 2018 ( α − β − γ ) 2 ( α 6 β 7 γ 7 + α 7 β 6 γ 7 + α 7 β 7 γ 6 ) + ( α 8 β 6 γ 6 + α 6 β 8 γ 6 + α 6 β 6 γ 8 ) 8 \dfrac{2016(\alpha^3-\beta^3-\gamma^3)+2017(\alpha^2-\beta^2-\gamma^2)+2018(\alpha-\beta-\gamma)}{\sqrt[8]{2(\alpha^{6}\beta^{7}\gamma^{7}+\alpha^{7}\beta^{6}\gamma^{7}+\alpha^{7}\beta^{7}\gamma^{6})+(\alpha^{8}\beta^{6}\gamma^{6}+\alpha^{6}\beta^{8}\gamma^{6}+\alpha^{6}\beta^{6}\gamma^{8})}}

= − 2016 + 2016 + 2016 ( α β γ ) 6 ( 2 α β + 2 β γ + 2 γ α + α 2 + β 2 + γ 2 ) 8 = \dfrac{-2016+2016+2016}{\sqrt[8]{(\alpha\beta\gamma)^6(2\alpha\beta+2\beta\gamma+2\gamma\alpha+\alpha^2+\beta^2+\gamma^2)}}

= 2016 ( − 2016 2017 ) 6 ( α + β + γ ) 2 8 = \dfrac{2016}{\sqrt[8]{(-\frac{2016}{2017})^6(\alpha+\beta+\gamma)^2}}

= 2016 ( − 2016 2017 ) 8 8 = \dfrac{2016}{\sqrt[8]{(-\frac{2016}{2017})^8}}

= 2016 2016 2017 = \dfrac{2016}{\frac{2016}{2017}}

= 2017 = 2017

7 Helpful 0 Interesting 0 Brilliant 0 Confused

I think you are saying that 2016 α 3 + 2017 α 2 + 2018 α 2016 \alpha^3 + 2017 \alpha^2 + 2018 \alpha is equal to − 2016 -2016 , and the same for β \beta and γ \gamma . But the correct equation, as you state above, is 2017 α 3 + 2016 α 2 + 2018 α = − 2016. 2017 \alpha^3 + 2016 \alpha^2 + 2018 \alpha = -2016.

Jon Haussmann - 4 years, 11 months ago

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