Of Course Vieta's formula, but...

Algebra Level 5

f ( x ) = x 3 − 2 x 2 + 6 x − 1 = 0 f(x)=x^3-2x^2+6x-1=0

The three roots of the above equation are α , β , γ \alpha,\beta,\gamma , then what is

α α 2 − α + 1 + β β 2 − β + 1 + γ γ 2 − γ + 1 = ? \frac{\alpha}{\alpha^2-\alpha+1}+\frac{\beta}{\beta^2-\beta+1}+\frac{\gamma}{\gamma^2-\gamma+1}=?


The answer is 0.25.

X X by X X , 17, Taiwan

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2 solutions

Mark Hennings
Aug 19, 2018

Note that X 3 − 2 X 2 + 6 X − 1 = ( X − 1 ) ( X 2 − X + 1 ) + 4 X X^3 - 2X^2 + 6X - 1 \; = \; (X-1)(X^2 - X + 1) + 4X so that α α 2 − α + 1 = 1 4 ( 1 − α ) \frac{\alpha}{\alpha^2 -\alpha + 1} \; = \; \tfrac14(1-\alpha) and similarly for β , γ \beta,\gamma . Thus we need to calculate 1 4 ( 3 − ∑ α ) = 1 4 ( 3 − 2 ) = 1 4 \tfrac14\left(3 - \sum \alpha\right) \; = \; \tfrac14(3-2) = \boxed{\tfrac14}

20 Helpful 1 Interesting 4 Brilliant 0 Confused

That's clever!

David Vreken - 2 years, 9 months ago

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Since α \alpha satisfies a cubic, any rational function of α \alpha must be equal to some quadratic function of α \alpha . We got lucky with a linear expression this time...

Mark Hennings - 2 years, 9 months ago
X X
Aug 19, 2018

α + β + γ = 2 , α β + β γ + γ α = 6 , α β γ = 1 \alpha+\beta+\gamma=2,\alpha\beta+\beta\gamma+\gamma\alpha=6,\alpha\beta\gamma=1

α α 2 − α + 1 + β β 2 − β + 1 + γ γ 2 − γ + 1 \space\space\space\space\dfrac{\alpha}{\alpha^2-\alpha+1}+\dfrac{\beta}{\beta^2-\beta+1}+\dfrac{\gamma}{\gamma^2-\gamma+1}

= 1 α − 1 + 1 α + 1 β − 1 + 1 β + 1 γ − 1 + 1 γ =\dfrac1{\alpha-1+\frac1{\alpha}}+\dfrac1{\beta-1+\frac1{\beta}}+\dfrac1{\gamma-1+\frac1{\gamma}}

= 1 2 − β − γ − 1 + β γ + 1 2 − γ − α − 1 + γ α + 1 2 − α − β − 1 + α β =\dfrac1{2-\beta-\gamma-1+\beta\gamma}+\dfrac1{2-\gamma-\alpha-1+\gamma\alpha}+\dfrac1{2-\alpha-\beta-1+\alpha\beta}

= 1 ( 1 − β ) ( 1 − γ ) + 1 ( 1 − γ ) ( 1 − α ) + 1 ( 1 − α ) ( 1 − β ) =\dfrac1{(1-\beta)(1-\gamma)}+\dfrac1{(1-\gamma)(1-\alpha)}+\dfrac1{(1-\alpha)(1-\beta)}

= 3 − ( α + β + γ ) ( 1 − α ) ( 1 − β ) ( 1 − γ ) =\dfrac{3-(\alpha+\beta+\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)}

= 1 f ( 1 ) = 0.25 =\dfrac1{f(1)}=0.25

11 Helpful 0 Interesting 2 Brilliant 0 Confused

@ricky Thegreat I was really surprised that you liked my problem after the like function was removed (8/27). How did you do that? (Or maybe the notifications went wrong so I recieved the mail late?)

X X - 2 years, 9 months ago

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