An algebra problem by Soumya Shrivastva

Algebra Level 4

If α , β , γ \alpha, \beta,\gamma are the roots of the equation x 3 − 3 x 2 + 3 x + 7 = 0 x^3-3x^2+3x+7 = 0 and ω \omega is the cube root of unity, compute

α − 1 β − 1 + β − 1 γ − 1 + γ − 1 α − 1 . \dfrac{\alpha - 1}{\beta - 1} + \dfrac{\beta- 1}{\gamma - 1} + \dfrac{\gamma - 1}{\alpha - 1} .

3 ω { 3\omega } 2 ω 3 2{ \omega }^{ 3 } 3 ω 2 3{ \omega }^{ 2 } Both 3 ω 2 3{ \omega }^{ 2 } and 3 ω 3\omega are correct

Soumya Shrivastva by Soumya Shrivastva , 20, India

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1 solution

Soumya Shrivastva
Jan 13, 2016

x 3 − 3 x 2 + 3 x + 7 − 8 = − 8 x 3 − 3 x 2 + 3 x − 1 = − 8 ( x − 1 ) 3 = − 8 x − 1 − 2 = 1 3 = 1 , ω , ω 2 α = − 1 , β = 1 − 2 ω , γ = 1 − 2 ω 2 α − 1 β − 1 + β − 1 γ − 1 + γ − 1 α − 1 = − 2 − 2 ω + − 2 ω − 2 ω 2 + − 2 ω 2 − 2 = 3 ω = 3 ω 2 { x }^{ 3 }-3{ x }^{ 2 }+3x+7-8=-8\\ { x }^{ 3 }-3{ x }^{ 2 }+3x-1=-8\\ { \left( x-1 \right) }^{ 3 }=-8\\ \frac { x-1 }{ -2 } =\sqrt [ 3 ]{ 1 } =1,\omega ,{ \omega }^{ 2 }\\ \alpha =-1,\quad \beta =1-2\omega ,\quad \gamma =1-2{ \omega }^{ 2 }\\ \frac { \alpha -1 }{ \beta -1 } +\frac { \beta -1 }{ \gamma -1 } +\frac { \gamma -1 }{ \alpha -1 } =\frac { -2 }{ -2\omega } +\frac { -2\omega }{ -2{ \omega }^{ 2 } } +\frac { -2{ \omega }^{ 2 } }{ -2 } =\frac { 3 }{ \omega } =3{ \omega }^{ 2 }\\

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