Roots of equation $x^3-x-1=0$

Question

If

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are the roots of equation $x^3 -x -1 =0$ then

$$ \frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma} $$

My attempt is in the attachment

I got answer $=0$ but in book answer is given as $-7$ . Where I do mistake by solving the question ?

Componendo-dividendo is not applied correctly. Verify that this step is incorrect by expanding both sides. — Sarvesh Ravichandran Iyer, Feb 26 '19 at 13:27
just a formatting tip: \alpha, \beta and \gamma gives the corresponding greek letters when in a math enviroment, that is surrounded by sollar signs ;) — Vinyl_cape_jawa, Feb 26 '19 at 13:32
Solving for $x$ from $t=(1+x)/(1-x)$ gives $x=(t-1)/(t+1)$. Therefore $(1+\alpha)/(1-\alpha)$ and friends are zeros of $$ \left(\frac{x-1}{x+1}\right)^3-\left(\frac{x-1}{x+1}\right)-1=\frac{-x^3-7x^2+x-1}{(1+x)^3}=0.$$ — Jyrki Lahtonen, Feb 26 '19 at 13:34
We don't apply CD and instead focus on the polynomial whose roots are the three quantities whose sum you are looking for. Now use Jyrki's hint to find this polynomial. Since you know the answer the polynomial should be clear from Jyrki's working — Sarvesh Ravichandran Iyer, Feb 26 '19 at 13:36
@jaykri can We solve this question without using transformation of equation concept ? — Tips, Feb 26 '19 at 13:38
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Please don't be discouraged by using pictures in questions. They are nice ways to provide illustrations to questions/answers. However, they are not a substitute to typing text/math. For some basic information about writing math at this site see e.g. basic help on mathjax notation, mathjax tutorial and quick reference. — GNUSupporter 8964民主女神地下教會, Feb 26 '19 at 14:52
https://math.stackexchange.com/questions/2157643/how-can-i-solve-the-equation-x3-x-1-0/2157645#2157645 and https://math.stackexchange.com/questions/2203364/solve-the-following-equation-x3-3x-sqrtx2 — lab bhattacharjee, Feb 26 '19 at 14:52

David Quinn · Accepted Answer · 2019-02-26T13:49:53.720

2

hint...write $$y=\frac{1+x}{1-x}\implies x=\frac{y-1}{y+1}$$ and substitute into the polynomial. Simplify the polynomial in $y$ and find the sum of the roots.

Alternatively,$$\Sigma\frac{1+\alpha}{1-\alpha}=-3+2\Sigma\frac{1}{1-\alpha}$$ $$=-3+2\frac{3+\Sigma\alpha\beta-2\Sigma\alpha}{1-\Sigma\alpha+\Sigma\alpha\beta-\alpha\beta\gamma}$$ $$=-3+2\frac{3-1-0}{1-0-1-1}=-7$$

edited Feb 26 '19 at 13:49

answered Feb 26 '19 at 13:36

David Quinn

35,087

DXT · Answer 2 · 2019-02-26T14:31:16.610

1

$$\sum \frac{1+\alpha}{1-\alpha}=2\sum\frac{1}{1-\alpha}-3\cdots (1)$$

Where $$\sum\frac{1+\alpha}{1-\alpha}=\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1-\gamma}{1-\gamma}.$$

Now adding $1$ in each term and

subtracting $-3$,We get equation no $(1)$

Now $$x^3-x-1=(x-\alpha)(x-\beta)(x-\gamma)$$

Taking $\ln$ and Differentiate both side and put $x=1$

$$\frac{3(1)^2-1}{(1)^3-(1)-1}=\frac{1}{1-\alpha}+\frac{1}{1-\beta}+\frac{1}{1-\gamma}$$

So we get $\displaystyle \sum \frac{1}{1-\alpha}=-2$

put into $(1)$, we have $$\displaystyle \frac{1+\alpha}{1-\alpha}=-4-3=-7.$$

edited Feb 26 '19 at 14:31

answered Feb 26 '19 at 13:40

DXT

12,047

Can you explain the first step.. – Tips Feb 26 '19 at 13:44
$\displaystyle \sum \frac{1+\alpha}{1-\alpha}=\sum\bigg(\frac{1+\alpha}{1-\alpha}+1\bigg)-3=2\sum\frac{1}{1-\alpha}-3$ – DXT Feb 26 '19 at 13:46
Sir , why you have added 1 then also subtracted 3 ?? – Tips Feb 26 '19 at 13:53
because $\displaystyle \sum \frac{1+\alpha}{1-\alpha}=\bigg(\frac{1+\alpha}{1-\alpha}+1\bigg)+\bigg(\frac{1+\beta}{1-\beta}+1\bigg)+\bigg(\frac{1+\gamma}{1-\gamma}+1\bigg)-3$ – DXT Feb 26 '19 at 13:55

Roots of equation $x^3-x-1=0$

2 Answers2