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Algebra Level 4

∑ α , β , γ 1 + α 1 − α \large \displaystyle \sum_{\alpha,\beta,\gamma} \frac{1+\alpha}{1-\alpha}

Given that α , β \alpha,\beta and γ \gamma are the roots of x 3 − x − 1 = 0 x^3-x-1=0 , then find the value of above expression.

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The answer is -7.

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Puneet Pinku by Puneet Pinku , 20, India

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3 solutions

Otto Bretscher
May 21, 2016

We substitute x = u + 1 x=u+1 and let a = α − 1 , b = β − 1 , c = γ − 1 a=\alpha -1, b=\beta-1,c=\gamma-1 be the roots of u 3 + 3 u 2 + 2 u − 1 u^3+3u^2+2u-1 . By Viète, the sum we seek is

− a + 2 a − b + 2 b − c + 2 c = − 3 − 2 a − 2 b − 2 c = − 3 − 2 b c + 2 a c + 2 a b a b c = − 3 − 2 × 2 1 = − 7 -\frac{a+2}{a}-\frac{b+2}{b}-\frac{c+2}{c}=-3-\frac{2}{a}-\frac{2}{b}-\frac{2}{c}=-3-\frac{2bc+2ac+2ab}{abc}=-3-\frac{2\times 2}{1}=\boxed{-7}

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Moderator note:

Great solution.

The substitution u = 1 + x 1 − x u = \frac{1+x}{1-x} also works.

I solved this question before attempting the problem here, so I only needed 10 seconds to solve this problem.

Hung Woei Neoh - 5 years ago

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Yes, I noticed that there have been a lot of questions about the roots of x 3 − x − 1 x^3-x-1 lately. I decided to solve just one of them, realizing that the others then become trivial, as you say.

Otto Bretscher - 5 years ago
Rohit Udaiwal
May 21, 2016

Note that by Vieta's Formuale for a cubic equation of the form a x 3 + b x 2 + c x + d ax^3+bx^2+cx+d has it roots α \alpha , β \beta and γ \gamma related to the coefficients of the equation by the relations:

  • α + β + γ = − b a \alpha+\beta+\gamma=\large \frac{-b}{a}
  • α β + β γ + γ α = c a \alpha\beta+\beta\gamma+\gamma\alpha=\large \frac{c}{a}
  • α β γ = − d a \alpha\beta\gamma=\large \frac{-d}{a}

Now we have ∑ α , β , γ 1 + α 1 − α = 1 + α 1 − α + 1 + β 1 − β + 1 + γ 1 − γ = ( 1 + α ) ( 1 − β ) ( 1 − γ ) + ( 1 + β ) ( 1 − γ ) ( 1 − α ) + ( 1 + γ ) ( 1 − α ) ( 1 − β ) ( 1 + α ) ( 1 + β ) ( 1 + γ ) = 3 − ( α + β + γ ) − ( α β + β γ + γ α ) + 3 α β γ 1 − ( α β γ ) + ( α β + β γ + γ α ) − α β γ [ Simplifying ! ] = 3 + b a − c a − 3 d a 1 + b a + c a + d a \begin{aligned} \displaystyle \sum_{\alpha,\beta,\gamma} \frac{1+\alpha}{1-\alpha} & =\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma} \\ & =\frac{(1+\alpha)(1-\beta)(1-\gamma)+(1+\beta)(1-\gamma)(1-\alpha)+(1+\gamma)(1-\alpha)(1-\beta)}{(1+\alpha)(1+\beta)(1+\gamma)} \\ & = \frac{3-(\alpha+\beta+\gamma)-(\alpha\beta+\beta\gamma+\gamma\alpha)+3\alpha\beta\gamma}{1-(\alpha\beta\gamma)+(\alpha\beta+\beta\gamma+\gamma\alpha)-\alpha\beta\gamma} \quad \quad [\text{Simplifying !}] \\ & =\frac{3+\frac{b}{a}-\frac{c}{a}-\frac{3d}{a}}{1+\frac{b}{a}+\frac{c}{a}+\frac{d}{a}} \end{aligned}

Now all we have to do is plug in the values of the coefficients of our equation x 3 − x − 1 = 0 x^3-x-1=0 ,where,clearly, a = 1 , b = 0 , c = − 1 a=1,b=0,c=-1 and d = − 1 d=-1 . ∑ α , β , γ 1 + α 1 − α = 3 + 3 + 1 1 − 1 − 1 = − 7 \large\begin{aligned} \displaystyle \sum_{\alpha,\beta,\gamma} \frac{1+\alpha}{1-\alpha} & =\frac{3+3+1}{1-1-1} \\ & = \large \displaystyle \boxed{\color{#20A900}{\color{#3D99F6}{-7}}} \end{aligned}

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Nice solution! I solved it in the same way while I was trying inspiration promblem by Akshat Sharda. I couldn't solve it but my mistake gave rise to new problems.. This can be also solved by transformation of equation according to their required roots. Check out the inspiration problem for thiskind of solution.

Puneet Pinku - 5 years ago

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Thanks :)Yes it often happens.Afterall Mistakes Give Rise To Problems !

Rohit Udaiwal - 5 years ago

Yes,I can see such a solution by Otto sir now.

Rohit Udaiwal - 5 years ago
Hung Woei Neoh
Jun 18, 2016

∑ n = { α , β , γ } 1 + n 1 − n = 1 + α 1 − α + 1 + β 1 − β + 1 + γ 1 − γ \displaystyle \sum_{n=\{\alpha,\beta,\gamma\}} \dfrac{1+n}{1-n}\\ =\dfrac{1+\alpha}{1-\alpha}+\dfrac{1+\beta}{1-\beta}+\dfrac{1+\gamma}{1-\gamma}

From Vieta's formula, we know that

α + β + γ = 0 α β + α γ + β γ = − 1 α β γ = 1 \alpha + \beta + \gamma = 0\\ \alpha\beta+\alpha\gamma + \beta\gamma = -1\\ \alpha\beta\gamma = 1

Now, to simplify our calculations, we substitute α = a + 1 \alpha = a + 1 , β = b + 1 \beta = b+1 and γ = c + 1 \gamma = c+1 into the expression and the equations

The expression becomes:

a + 2 − a + b + 2 − b + c + 2 − c = − ( b c ( a + 2 ) + a c ( b + 2 ) + a b ( c + 2 ) a b c ) = − ( 3 a b c + 2 ( a b + a c + b c ) a b c ) \dfrac{a+2}{-a} + \dfrac{b+2}{-b} + \dfrac{c+2}{-c}\\ =-\left(\dfrac{bc(a+2) + ac(b+2) + ab(c+2)}{abc}\right)\\ =-\left(\dfrac{3abc+2(ab+ac+bc)}{abc} \right)

The equations become (simplify yourself, I'm too lazy to type out all the steps):

a + 1 + b + 1 + c + 1 = 0 ⟹ a + b + c = − 3 ( a + 1 ) ( b + 1 ) + ( a + 1 ) ( c + 1 ) + ( b + 1 ) ( c + 1 ) = − 1 ⟹ a b + a c + b c = 2 ( a + 1 ) ( b + 1 ) ( c + 1 ) = 1 ⟹ a b c = 1 a + 1 + b + 1 + c + 1 = 0 \implies a+b+c = -3\\ (a+1)(b+1) + (a+1)(c+1) + (b+1)(c+1) = -1 \implies ab+ac+bc = 2\\ (a+1)(b+1)(c+1) = 1 \implies abc=1

Substitute these values into the expression to get the answer:

− ( 3 a b c + 2 ( a b + a c + b c ) a b c ) = − ( 3 ( 1 ) + 2 ( 2 ) 1 ) = − 7 -\left(\dfrac{3abc+2(ab+ac+bc)}{abc} \right)\\ =-\left(\dfrac{3(1)+2(2)}{1}\right)\\ =\boxed{-7}

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