Roots of third degree equation

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Roots of third degree equation

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Source: Canada 1996

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#1 h Mar 4, 2006, 12:14 PM • 3 Y

Y by Adventure10, Mango247, yshk

If $\alpha$ , $\beta$ , and $\gamma$ are the roots of $x^3 - x - 1 = 0$ , compute $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$ .

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metru

#2 h Mar 4, 2006, 12:45 PM • 10 Y

Y by NewBeginning, bel.jad5, yousseframzi, Adventure10, RedFireTruck, AlexCenteno2007, persamaankuadrat, EtacticToe, and 2 other users

If $\alpha$ , $\beta$ , and $\gamma$ are the roots of $x^3 - x - 1 = 0$ ,
compute $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$ .

(Canada National Olympiad 1996)

Solution:

If $\alpha ,\beta$ and $\gamma$ are the roots of $x^3 - x - 1 = 0$ ,
then $\frac{{1 + \alpha }}{{1 - \alpha }},\frac{{1 + \beta }}{{1 - \beta }}$ and $\frac{{1 + \gamma }}{{1 - \gamma }}$ are the roots of
$\left( {\frac{{u - 1}} {{u + 1}}} \right)^3 - \left( {\frac{{u - 1}} {{u + 1}}} \right) - 1 = 0 \Leftrightarrow$

$\Leftrightarrow u^3 + 7u^2 - u + 1 = 0$

Then we have:
$\frac{{1 + \alpha }} {{1 - \alpha }} + \frac{{1 + \beta }} {{1 - \beta }} + \frac{{1 + \gamma }} {{1 - \gamma }} = - 7$

This post has been edited 2 times. Last edited by metru, Mar 4, 2006, 4:39 PM

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#3 h Mar 4, 2006, 4:21 PM • 1 Y

Y by Adventure10

Clever, Metru!

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#4 h Apr 10, 2006, 2:32 PM • 2 Y

Y by Adventure10, Mango247

metru wrote:

[If $\alpha ,\beta$ and $\gamma$ are the roots of $x^3 - x - 1 = 0$ ,
then $\frac{{1 + \alpha }}{{1 - \alpha }},\frac{{1 + \beta }}{{1 - \beta }}$ and $\frac{{1 + \gamma }}{{1 - \gamma }}$ are the roots of
$\left( {\frac{{u - 1}} {{u + 1}}} \right)^3 - \left( {\frac{{u - 1}} {{u + 1}}} \right) - 1 = 0$

Why?

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#5 h Apr 10, 2006, 4:05 PM • 4 Y

Y by Adventure10, Mango247, Mango247, Mango247

If $u = \frac{1+\alpha}{1-\alpha}$ , then $u-u\alpha=1+\alpha\Longleftrightarrow \alpha\cdot (1+u)=u-1$ , so $\alpha = \frac{u-1}{u+1}$ . Since $\alpha^3-\alpha-1=0$ , we must have $\left(\frac{u-1}{u+1}\right)^3-\left(\frac{u-1}{u+1}\right)-1=0.$

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shyong

#6 h Apr 13, 2006, 10:15 AM • 2 Y

Y by Adventure10, Mango247

here is my friend's solution .

Letting $S=\sum\frac{1+\alpha}{1-\alpha}$ , then $S+3=\sum\frac{2}{1-\alpha}=2\frac{\sum (1-\beta)(1-\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)}$

Since $f(x)=x^3-x-1=(x-\alpha)(x-\beta)(x-\gamma)$ , $f(1)=(1-\alpha)(1-\beta)(1-\gamma)=-1$

So the denominator of $S+3$ is $-1$ while the numerator (after expanding and using vieta) we find that it is $4$ .So $S+3=-4$ $\implies S=-7$

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#7 h Oct 16, 2006, 1:41 AM • 2 Y

Y by Adventure10, Mango247

shyong wrote:

here is my friend's solution .

Letting $S=\sum\frac{1+\alpha}{1-\alpha}$ , then $S+3=\sum\frac{2}{1-\alpha}=2\frac{\sum (1-\beta)(1-\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)}$

Since $f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$ , $f(1)=(1-\alpha)(1-\beta)(1-\gamma)=-1$

So the denominator of $S+3$ is $-1$ while the numerator (after expanding and using vieta) we find that it is $4$ .So $S+3=-4$ $\implies S=-7$

Can some body please explain how
$f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$

Just because \alpha , \beta , \gamma are roots of f(x) doesn't mean above is true.

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#8 h Nov 22, 2013, 10:34 AM • 1 Y

Y by Adventure10

Quote:

Can some body please explain how
$f(x)=x^{3}-x-1=(x-\alpha)(x-\beta)(x-\gamma)$

Just because \alpha , \beta , \gamma are roots of f(x) doesn't mean above is true.

This is just euclidean division on polynomials , $P(\alpha) = 0$ means $x-\alpha$ divides $P(x)$ i.e , there exists a polynomial $Q$ (of degree $2$ ) such that $P(x) = (x-\alpha) Q(x)$ for all $X$ (This is because the ring of polynomials is euclidean just like $\mathbb{Z}$ Do the same thing , since $\beta$ is another root of $P$ We have necessarily $Q(\beta)=0$ (since $\beta \ne \alpha$ ) and so on .

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#9 h Nov 23, 2017, 6:24 PM • 1 Y

Y by Adventure10

Note that $\sum\limits_{sym}^{} \frac{1+\alpha}{1-\alpha} = \sum\limits_{sym}^{}\frac{1-\alpha + 2\alpha}{1-\alpha} = \sum\limits_{sym}^{}1+\frac{2\alpha}{1-\alpha} = 3+ 2\sum\limits_{sym}^{}\frac{\alpha}{1-\alpha}$ .

We can rewrite the equation $x^3-x-1=0$ as $x=x^3-1$ . Hence, $3+2\sum\limits_{sym}^{}\frac{\alpha}{1-\alpha} = 3+2\sum\limits_{sym}^{}\frac{\alpha^3-1}{1-\alpha} = 3-2\sum\limits_{sym}^{} \frac{\alpha^3-1}{\alpha-1}$

Since $1$ is not a root of the equation, we can divide and get

$3-2\sum\limits_{sym}^{} \alpha^2+\alpha+1 = 3-2(s^2-2(\alpha\beta + \beta\gamma + \gamma\alpha) + s + 3) = 3-2(0^2-2(-1) + 0 +3) = -7$

@below thanks!

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#10 h Nov 24, 2017, 11:33 AM • 2 Y

Y by Adventure10, Mango247

NewBeginning wrote:

$3-2\sum\limits_{sym}^{} \alpha^2+\alpha+1 = 3-2(s^2-2(\alpha\beta + \beta\gamma + \gamma\alpha) + s + 1)$

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

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#11 h Apr 24, 2018, 11:29 AM • 2 Y

Y by Adventure10, Mango247

Isn't this simply vieta?

$Solution:$
Firstly let $P(x) = x^3 - x- 1$ . Then $\alpha,\beta,\gamma$ are the roots of $P(x)$ . For brevity, let $\alpha =a , \beta =b, \gamma=c$ .(Lazinesss

)
Then by vieta's theorem, we have that:
$\longrightarrow \sum_{cyc} a = 0$

$\longrightarrow \sum_{cyc} ab = -1$

$\longrightarrow abc = 1$ .

Expanding the given equation we have:

$\dfrac{\sum_{cyc}(1+a)(1-b)(1-c)}{(1-a)(1-b)(1-c)}$

$\iff \dfrac{7}{-1}$

Hence the answer is $-7$

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#12 h Dec 19, 2022, 2:48 AM

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Let $P$ be the polynomial $x^3-x-1$ . Then $\sum_{\mathrm{cyc}} \frac{1+\alpha}{1-\alpha} = \frac{\sum_{\mathrm{cyc}} (1+\alpha)(1-\beta)(1-\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)} = \frac{3+3\alpha\beta\gamma - \sum \alpha - \sum \alpha \beta}{P(1)} = \boxed{-7}.$

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#13 h Jan 13, 2023, 5:08 PM

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Ans: -7
1+a÷1-a=t
=> 1÷a=t+1÷t-1=>a=t-1÷t+1
Putting the value of 'a' in given Equation, we get The sum of roots as -7

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#14 h Jan 13, 2023, 11:21 PM

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$\begin{align*} \frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma} &=\frac{(1+\alpha)(1-\beta)(1-\gamma)+(1-\alpha)(1+\beta)(1-\gamma)+(1-\alpha)(1-\beta)(1+\gamma)}{(1-\alpha)(1-\beta)(1-\gamma)}\\ &=\frac{3-(\alpha+\beta+\gamma)-(\alpha\beta+\alpha\gamma+\beta\gamma)+3\alpha\beta\gamma}{1-(\alpha+\beta+\gamma)+(\alpha\beta+\alpha\gamma+\beta\gamma)-\alpha\beta\gamma}\\ &=\frac{3-0+1+3\cdot 1}{1-0-1-1}\\ &=-7\\ \end{align*}$

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#15 h Apr 11, 2023, 8:41 PM

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Let $1-\alpha=a$ , $1-\beta=b$ , and $1-\gamma=c$ . Then $a$ , $b$ , and $c$ are the roots of $(1-x)^3-(1-x)-1=-x^3+3x^2-2x-1$ . Therefore, $\frac{2-a}{a}+\frac{2-b}{b}+\frac{2-c}{c}=2(\frac1a+\frac1b+\frac1c)-3=2(\frac{2}{-1})-3=-7$ .

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#16 h Apr 20, 2023, 2:53 AM

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solution

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#17 h Apr 20, 2023, 3:28 AM • 1 Y

Y by mbrioche

@cinnamon_e,
very nice job! I'm highly impressed. Oh...if you don't mind accepting my friend request....
Later

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#18 h Apr 20, 2023, 3:34 AM

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shobber wrote:

If $\alpha$ , $\beta$ , and $\gamma$ are the roots of $x^3 - x - 1 = 0$ , compute $S=\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$ .

Let $P(x)=x^3-x-1$ , then $\frac{P’(x)}{P(x)}=\sum\frac1{x-\alpha}$ , therefore
$S=-3+ 2\frac{P’(1)}{P(1)} =-7$

This post has been edited 1 time. Last edited by cazanova19921, Apr 20, 2023, 3:34 AM

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#19 h May 29, 2023, 9:47 AM

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By Vietas Relation, we get $\alpha+\beta+\gamma=0$ $\alpha\beta+\beta\gamma+\gamma\alpha=-1$ $\alpha\beta\gamma=1$ We need to find the value of $\sum_{cyc}{}\frac{1+\alpha}{1-\alpha}.$ Now we have to find an equation whose roots are $\frac{1+\alpha}{1-\alpha},$ $\frac{1+\beta}{1-\beta},$ $\frac{1+\gamma}{1-\gamma}$
Now, $\frac{1+\alpha}{1-\alpha}=t\implies\alpha=\frac{t-1}{t+1}$ Putting the value of $\alpha$ in given $eq^n$ we get $\left(\frac{t-1}{t+1}\right)^3-\left(\frac{t-1}{t+1}\right)-1=0.$ After simplify we get our reqd. equation as $t^3+7t^2-t+1=0.$ therefore $\sum_{cyc}{}\frac{1+\alpha}{1-\alpha}=-7$ Is the correct answer....
VIETAS RELATION :love:

:love:

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#20 h Jul 18, 2023, 1:59 PM

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I think solutions like #2 and #16 are actually slightly incomplete. It should be mentioned that the roots are distinct in order for the proof to actually work. Otherwise there is no guarantee that the multiplicities of the roots are preserved.

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#21 h Aug 5, 2023, 8:02 PM

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Taken from my notebook:
Clearly,
$\frac{1+\alpha}{1-\alpha}+\frac{1+\beta}{1-\beta}+\frac{1+\gamma}{1-\gamma}=3+2\left(\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}+\frac{\gamma}{1-\gamma}\right).$ We also know that
$\begin{align*} \alpha & =\alpha^3-1, \\ \beta & =\beta^3-1, \\ \gamma & =\gamma^3-1. \end{align*}$ After substituting, our desired result becomes,
$\begin{align*} 3+2\left(\frac{\alpha}{1-\alpha}+\frac{\beta}{1-\beta}+\frac{\gamma}{1-\gamma}\right) = & 3+2(-\alpha^2-\alpha-1-\beta^2-\beta-1-\gamma^2-\gamma-1). \end{align*}$ By Vieta's, we have
$\begin{align*} \alpha^2+\beta^2+\gamma^2 & = 2, \\ \alpha+\beta+\gamma & =0. \end{align*}$ Hence, our desired result becomes
$3+2(-\alpha^2-\alpha-1-\beta^2-\beta-1-\gamma^2-\gamma-1)=3+2(-5)=\boxed{-7}$

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Ianis

#22 h Aug 5, 2023, 8:57 PM

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Filipjack wrote:

I think solutions like #2 and #16 are actually slightly incomplete. It should be mentioned that the roots are distinct in order for the proof to actually work. Otherwise there is no guarantee that the multiplicities of the roots are preserved.

Anyway, that's easy. Right now I can think of two ways of proving it.

Elementary:
If $f(x)=x^3-x-1$ had a multiple root then it would be a root of $f'(x)=3x^2-1$ , but the roots of $3x^2-1$ are not roots of $x^3-x-1$ . Hence $\alpha \neq \beta \neq \gamma \neq \alpha$ .

Advanced:
By the Rational Root Theorem we have that the only possible rational roots of $f(x)=x^3-x-1$ are $\pm 1$ , but $f(1)=f(-1)=-1$ , so $f$ does not have rational roots and therefore it is irreducible in $\mathbb{Q}[x]$ . From $\operatorname{char}\mathbb{Q}=0$ we get that $f$ is separable over $\mathbb{Q}$ . Hence $\alpha \neq \beta \neq \gamma \neq \alpha$ .

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#23 h Aug 7, 2023, 9:14 AM

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Ianis wrote:

Filipjack wrote:

I think solutions like #2 and #16 are actually slightly incomplete. It should be mentioned that the roots are distinct in order for the proof to actually work. Otherwise there is no guarantee that the multiplicities of the roots are preserved.

Anyway, that's easy. Right now I can think of two ways of proving it.

Elementary:
If $f(x)=x^3-x-1$ had a multiple root then it would be a root of $f'(x)=3x^2-1$ , but the roots of $3x^2-1$ are not roots of $x^3-x-1$ . Hence $\alpha \neq \beta \neq \gamma \neq \alpha$ .

Advanced:
By the Rational Root Theorem we have that the only possible rational roots of $f(x)=x^3-x-1$ are $\pm 1$ , but $f(1)=f(-1)=-1$ , so $f$ does not have rational roots and therefore it is irreducible in $\mathbb{Q}[x]$ . From $\operatorname{char}\mathbb{Q}=0$ we get that $f$ is separable over $\mathbb{Q}$ . Hence $\alpha \neq \beta \neq \gamma \neq \alpha$ .

Well, yeah, it's easy, but I didn't make that remark because that step was hard, but rather to emphasize this common (and subtle) error: If $r_1, \ldots, r_n$ are the roots of some polynomial $p$ and I prove that

$r_i$ is a root of $p$ $\implies$ $\varphi(r_i)$ is a root of $q,$ for $i=\overline{1,n},$

this won't actually prove that the multiplicity is preserved. If $r_1$ is a double root, for instance, the above implications merely say twice that $\varphi(r_1)$ is a root of $q,$ but they don't actually say that $\varphi(r_i)$ is a double root of $q.$

Anyway, we don't really need the roots to be distinct. That was just a quick fix. Using the factorization of $p,$ we can easily see that $p\left(\tfrac{X-b}{a}\right)$ has the roots $ar_1+b,\ldots,ar_n+b$ (with multiplicities preserved) and $X^np \left(\tfrac{1}{X} \right)$ has the roots $\tfrac{1}{r_1},\ldots\tfrac{1}{r_n}$ (again, with multiplicities preserved; here we assume that the roots are non-zero), so we get the result if we use the right compositions.

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#24 h Aug 7, 2023, 9:42 AM • 1 Y

Y by Rounak_iitr

By Vieta's formula, we get $\left\{ \begin{array}{l} \alpha + \beta + \gamma = 0\\ \alpha \beta + \beta \gamma + \gamma \alpha = - 1\\ \alpha \beta \gamma = 1 \end{array} \right. ,$ then we have $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}=\frac{{3 - (\alpha + \beta + \gamma ) - (\alpha \beta + \beta \gamma + \gamma \alpha ) + 3\alpha \beta \gamma }}{{1 - (\alpha + \beta + \gamma ) + (\alpha \beta + \beta \gamma + \gamma \alpha ) - \alpha \beta \gamma }}=-7.$

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#25 h Aug 28, 2023, 2:19 AM

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By Vieta's formula
$\begin{align*} \alpha\beta\gamma = 1 \\ \alpha\beta + \beta\gamma + \alpha\gamma = -1 \\ \alpha + \beta + \gamma = 0 \end{align*}$
and expressing $\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma}$ with a common denominator we get

$\frac{3+3\alpha\beta\gamma-\alpha\gamma-\beta\gamma-\alpha\beta-\alpha-\gamma-\beta}{1-\alpha-\beta-\gamma+\alpha\beta+\alpha\gamma+\gamma\beta-\alpha\beta\gamma} = \frac{3+3+1}{-1} = \boxed{-7}$

This post has been edited 1 time. Last edited by LostDreams, Aug 28, 2023, 2:20 AM
Reason: forgot to add $$

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#26 h Oct 13, 2023, 4:51 PM

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We have

$\begin{align*} &\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma} \\ = \ &\frac{1+\alpha}{1-\alpha} + 1 + \frac{1+\beta}{1-\beta} + 1 + \frac{1+\gamma}{1-\gamma} + 1 - 3 \\ = \ & -2\left(\frac{1}{\alpha-1} + \frac{1}{\beta-1} + \frac{1}{\gamma-1} \right)-3 \end{align*}$
The polynomial with roots $\alpha-1$ , $\beta-1$ , and $\gamma-1$ is $(x+1)^3-(x+1)-1=x^3 + 3 x^2 + 2 x - 1$ . Then, the polynomial with roots $\frac{1}{\alpha-1}$ , $\frac{1}{\beta-1}$ , $\frac{1}{\gamma-1}$ is $-x^3+2x^2+3x+1$ , so our answer is

$-2\left(\frac{1}{\alpha-1} + \frac{1}{\beta-1} + \frac{1}{\gamma-1} \right)-3 = -2(2)-3 = \boxed{-7}.$

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#27 h Dec 1, 2023, 11:20 AM

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By Vieta's formula, we know that $u=a+b+c=0$ , $v=ab+bc+ca=-1$ and $w=abc=1$ .
And since the desired sum can be rewrited as
$\dfrac{3+3w-v-u}{1+v-u-w}$ we find that $S=-7$

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#28 h Dec 3, 2023, 9:55 AM

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If $\alpha, \beta$ & $\gamma$ are the zeroes of $f(x) =ax^3+bx^2+cx+d,$ then $\frac{k+\alpha}{k-\alpha}, \frac{k+\beta}{k-\beta}$ and $\frac{k+\gamma}{k-\gamma}$ are the zeroes of $g(x) = (ak^3+bk^2+ck+d)x^3-(3ak^3+bk^2-ck-3d)x^2+(3ak^3-bk^2-ck+3d)x-(ak^3-bk^2+ck-d).$ $\boxed{\therefore \frac{k+\alpha}{k-\alpha}+\frac{k+\beta}{k-\beta}+\frac{k+\gamma}{k-\gamma}=\frac{3ak^3+bk^2-ck-3d}{ak^3+bk^2+ck+d}}$

This post has been edited 5 times. Last edited by Entrepreneur, Dec 3, 2023, 7:08 PM

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#29 h Dec 17, 2023, 3:45 PM

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Let $r$ be a root of $f(x)=x^3-x-1$ . Through addition, we have that
$\frac{1+r}{1-r}=-1+\frac{2}{1-r},$ however, since $r$ is a root of $f(x)$ , we have that
$r^3-r-1=0 \iff r^3-r=1 \iff -r^2-r=\frac{1}{1-r} \iff -2r^2-2r=\frac{2}{1-r},$ meaning that
$\frac{1+r}{1-r}=-1-2r-2r^2,$ which gives that
$\frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma} = -3-2(\alpha+\beta+\gamma)-2(\alpha^2+\beta^2+\gamma^2).$ By Vieta's formulas, we have that $\alpha+\beta+\gamma=0$ and $\alpha^2+\beta^2+\gamma^2=0^2-2(-1)=2$ , giving us an answer of $-3-2(0)-2(2),$ or $-7$ .

This post has been edited 1 time. Last edited by peppapig_, Dec 17, 2023, 3:46 PM
Reason: Negative sign

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KOFSCIMQOMISART

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KOFSCIMQOMISART

#30 h Jan 9, 2024, 3:52 PM

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Let a,b,c be the roots, so we have that a+b+c=0, abc=1 and ab+bc+ac=-1
Let H be our expression so H=7/(1-a)(1-b)(1-c)
It is easy to check that (1-a)(1-b)(1-c)=-1 and finally we get H=-1

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#31 h Apr 16, 2024, 10:33 PM

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Solution

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Markas

#32 h May 5, 2024, 1:24 PM

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We want to find A = $\frac{1 + \alpha}{1 - \alpha} + \frac{1 + \beta}{1 - \beta} + \frac{1 + \gamma}{1 - \gamma} = \frac{(1 + \alpha)(1 - \beta)(1 - \gamma) + (1 + \beta)(1 - \alpha)(1 - \gamma) + (1 + \gamma)(1 - \alpha)(1 - \beta)}{(1 - \alpha)(1 - \beta)(1 - \gamma)} = \frac{3 + 3\alpha\beta\gamma - (\alpha + \beta + \gamma) - (\alpha\beta + \alpha\gamma + \beta\gamma)}{1 - (\alpha + \beta + \gamma) + (\alpha\beta + \alpha\gamma + \beta\gamma) - \alpha\beta\gamma}$ . Now from Vieta's we get that $\alpha + \beta + \gamma = 0$ , $\alpha\beta + \alpha\gamma + \beta\gamma = -1$ , $\alpha\beta\gamma = 1$ . Now we plug in these sums in the equation for A we got $\Rightarrow$ $A = \frac{3 + 3.1 - 0 - (-1)}{1 - 0 + (-1) - 1} = \frac{7}{-1} = -7$ $\Rightarrow$ the answer is -7.

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combowomborhombo

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combowomborhombo

#33 h Aug 18, 2024, 8:36 PM

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We can construct a new polynomial with roots:

$\frac{1 + \alpha}{1 - \alpha}, \quad \frac{1 + \beta}{1 - \beta}$
The manipulation involves the transformation:

$\left( \frac{u - 1}{u + 1} \right)^3 - \left( \frac{u - 1}{u + 1} \right) - 1 = 0$
This expression simplifies to the polynomial:

$u^3 + 7u^2 - u + 1 = 0$
Using Vieta's formulas, we can find the sum of the roots of this new polynomial, which is:

$\boxed{-7}$

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#34 h Nov 4, 2024, 4:49 PM

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Let $x$ be the desired expression. Note that $x+3=2\bigg(\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}\bigg).$ We can create a common denominator for $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c}$ to find that it is $-2$ . The answer is $\boxed{-7}$ .

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#35 h Apr 25, 2025, 3:37 PM

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Let $P(x) = x^3 - x - 1.$ First, we have that $P(1) = (1 - \alpha)(1 - \beta)(1 - \gamma) = -1.$ We also have that from Vietas, $\alpha + \beta + \gamma = 0, \alpha\beta + \alpha\gamma + \beta\gamma = -1.$ Thus,
$\begin{align*} \frac{1+\alpha}{1-\alpha} + \frac{1+\beta}{1-\beta} + \frac{1+\gamma}{1-\gamma} & = -3 + \frac{2}{1-\alpha} + \frac{2}{1-\beta} + \frac{2}{1-\gamma} \\ & = -3 + 2 \left(\frac{3 - 2\alpha - 2\beta - 2\gamma + \alpha\beta + \alpha\gamma + \beta\gamma}{(1 - \alpha)(1 - \beta)(1 - \gamma)}\right) \\ & = -3 + 2 \left(\frac{3 - 1}{-1}\right) = \boxed{-7} \end{align*}$

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#36 h May 17, 2025, 4:30 PM

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We let the polynomial $Q(x) = (1-x)^3-(1-x)-1$ , which has roots $1-\alpha$ , $1-\beta$ , $1-\gamma$ . Expanding, we have that $Q(x) = -x^3+3x^2-2x-1.$ Now, let the roots of $Q(x)$ be $r$ , $s$ , and $t$ such that
$\begin{align*} r &= 1-\alpha \\ s &= 1-\beta \\ t &= 1-\gamma. \\ \end{align*}$ Then, we are solving for $\dfrac{2-r}{r}+\dfrac{2-s}{s}+\dfrac{2-t}{t} = \dfrac{2(rs+rt+st)}{rst}-3$ which clearly by Vieta's is equal to $\boxed{-7}$ .

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