Application of Equations #1

Algebra Level 3

Three complex numbers α , β , γ \alpha,\beta,\gamma satisfy below three equations.

  • α + β + γ = 1 \alpha+\beta+\gamma=1

  • α 2 + β 2 + γ 2 = 3 \alpha^2+\beta^2+\gamma^2=3

  • α β γ = 1 \alpha\beta\gamma=1

Find the value of α 10 − α 9 − α 8 + β 10 − β 9 − β 8 + γ 10 − γ 9 − γ 8 \alpha^{10}-\alpha^9-\alpha^8+\beta^{10}-\beta^9-\beta^8+\gamma^{10}-\gamma^9-\gamma^8 .


The answer is 71.

Boi (보이) by Boi (보이) , 19, South Korea

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1 solution

Boi (보이)
Jun 6, 2017

( α + β + γ ) 2 = α 2 + β 2 + γ 2 + 2 ( α β + β γ + γ α ) ; (\alpha+\beta+\gamma)^2=\alpha^2+\beta^2+\gamma^2+2(\alpha\beta+\beta\gamma+\gamma\alpha); .

α β + β γ + γ α = − 1 \alpha\beta+\beta\gamma+\gamma\alpha=-1 .

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One of the cubic equations that have α , β , γ \alpha, \beta, \gamma as their roots is ( x − α ) ( x − β ) ( x − γ ) = 0 ; (x-\alpha)(x-\beta)(x-\gamma)=0; .

x 3 − ( α + β + γ ) x 2 + ( α β + β γ + γ α ) x − α β γ = 0 ; x^3-(\alpha+\beta+\gamma)x^2+(\alpha\beta+\beta\gamma+\gamma\alpha)x-\alpha\beta\gamma=0;

x 3 − x 2 − x − 1 = 0 ; x^3-x^2-x-1=0;

x 3 = x 2 + x + 1 x^3=x^2+x+1 .

And since this equation has α , β , γ \alpha, \beta, \gamma as its roots,

α 3 = α 2 + α + 1 , β 3 = β 2 + β + 1 γ 3 = γ 2 + γ + 1 \alpha^3=\alpha^2+\alpha+1, \quad \beta^3=\beta^2+\beta+1\, \quad \gamma^3=\gamma^2+\gamma+1 .

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Multiply α n , β n , γ n \alpha^{n},\beta^{n},\gamma^{n} to each equation and define f ( n ) = α n + β n + γ n f(n)=\alpha^n+\beta^n+\gamma^n where n n is an integer that is not negative.

f ( n + 3 ) = f ( n + 2 ) + f ( n + 1 ) + f ( n ) f(n+3)=f(n+2)+f(n+1)+f(n) . Also, f ( n + 3 ) − f ( n + 2 ) − f ( n + 1 ) = f ( n ) f(n+3)-f(n+2)-f(n+1)=f(n) .

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Since f ( 0 ) = 3 , f ( 1 ) = 1 , f ( 2 ) = 3 f(0)=3, \quad f(1)=1,\quad f(2)=3 ,

f ( 3 ) = f ( 2 ) + f ( 1 ) + f ( 0 ) = 3 + 1 + 3 = 7 f ( 4 ) = f ( 3 ) + f ( 2 ) + f ( 1 ) = 7 + 3 + 1 = 11 f ( 5 ) = f ( 4 ) + f ( 3 ) + f ( 2 ) = 11 + 7 + 3 = 21 f ( 6 ) = f ( 5 ) + f ( 4 ) + f ( 3 ) = 21 + 11 + 7 = 39 f(3)=f(2)+f(1)+f(0)=3+1+3=7 \\ f(4)=f(3)+f(2)+f(1)=7+3+1=11 \\ f(5)=f(4)+f(3)+f(2)=11+7+3=21 \\ f(6)=f(5)+f(4)+f(3)=21+11+7= 39

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∴ α 10 − α 9 − α 8 + β 10 − β 9 − β 8 + γ 10 − γ 9 − γ 8 = f ( 10 ) − f ( 9 ) − f ( 8 ) = f ( 7 ) = f ( 6 ) + f ( 5 ) + f ( 4 ) = 39 + 21 + 11 = 71 \therefore \alpha^{10}-\alpha^9-\alpha^8+\beta^{10}-\beta^9-\beta^8+\gamma^{10}-\gamma^9-\gamma^8 \\ =f(10)-f(9)-f(8) \\ =f(7) \\ =f(6)+f(5)+f(4) \\ =39+21+11 \\ = \boxed{71}

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