MHB Evaluate (1-a)/(1+a)+(1-b)/(1+b)+(1-c)/(1+c)

[anemone]

If $a,\,b,\,c$ are the roots of $x^3-x-1=0$, evaluate $\dfrac{1-a}{1+a}+\dfrac{1-b}{1+b}+\dfrac{1-c}{1+c}$.

 

[kaliprasad]

Spoiler

$\dfrac{1-a}{1+a}=1- \dfrac{2a}{1+a}=1-\dfrac{2}{1+\dfrac{1}{a}}$
$a,\,b,\,c$ are the roots of $x^3-x-1=0$
so $\dfrac{1}{a},\,\dfrac{1}{b},\,\dfrac{1}{c}$ are the roots of $\dfrac{1}{x^3}-\dfrac{1}{x}-1=0$
or $x^3+x^2-1=0$
so $1+\dfrac{1}{a},\,1+\dfrac{1}{b},\,1+\dfrac{1}{c}$ are the roots of
$(x-1)^3+(x-1)^2-1=0$
or $x^3-3x^2+3x-1 +x^2-2x+1-1=0$
or $x^3-2x^2+x-1=0$
so $\dfrac{1}{1+\dfrac{1}{a}},\,\dfrac{1}{1+\dfrac{1}{b}},\,\dfrac{1}{1+\dfrac{1}{c}}$ are the roots of
$\dfrac{1}{x^3}-\dfrac{2}{x^2}+\dfrac{1}{x}-1=0$
or
$x^3-x^2+2x-1= 0$
so $\dfrac{1}{1+\dfrac{1}{a}}+\,\dfrac{1}{1+\dfrac{1}{b}}+\,\dfrac{1}{1+\dfrac{1}{c}}= 1$
or $\dfrac{a}{1+a}+\,\dfrac{b}{1+b}+\,\dfrac{c}{1+c}= 1$
or $\dfrac{2a}{1+a}+\,\dfrac{2b}{1+b}+\,\dfrac{2c}{1+c}= 2$
or $1- \dfrac{2a}{1+a}+1- \,\dfrac{2b}{1+b}+1-\,\dfrac{2c}{1+c}= 3-2$
or $\dfrac{1-a}{1+a}+\,\dfrac{1-b}{1+b}+\,\dfrac{1-c}{1+c}= 1$

 

[soroban]

Hello, anemone!

If $a,\,b,\,c$ are the roots of $x^3-x-1=0$,
evaluate $\,Y \;=\;\dfrac{1-a}{1+a}+\dfrac{1-b}{1+b}+\dfrac{1-c}{1+c}$

[sp]
Since $a,\,b,\,c$ are roots of the cubic: $\;\begin{Bmatrix}a+b+c &=& 0 \\ ab + bc + ac &=& \text{-}1 \\ abc &=& 1 \end{Bmatrix}$Then: $\,Y \,=\,\dfrac{(1-a)(1+b)(1+c) + (1+a)(1-b)(1+c) + (1+a)(1+b)(1-c)}{(1+a)(1+b)(1+c)} $

The numerator simplifies to:
$\quad 3 + (a+b+c) - (ab + bc + ac) - 3abc$
$\qquad =\:3 + 0 - (\text{-}1) - 3(1) \;=\;1$

The denominator simplifies to:
$\quad 1 + (a+b+c) + (ab+bc+ac) + abc$
$\qquad =\:1+0+(\text{-}1)+1 \;=\;1$

Therefore: $\:Y \;=\;\dfrac{1}{1} \;=\;1$
[/sp]

 

[kaliprasad]

Hello, anemone![sp]
Since $a,\,b,\,c$ are roots of the cubic: $\;\begin{Bmatrix}a+b+c &=& 0 \\ ab + bc + ac &=& \text{-}1 \\ abc &=& 1 \end{Bmatrix}$Then: $\,Y \,=\,\dfrac{(1-a)(1+b)(1+c) + (1+a)(1-b)(1+c) + (1+a)(1+b)(1-c)}{(1+a)(1+b)(1+c)} $

The numerator simplifies to:
$\quad 3 + (a+b+c) - (ab + bc + ac) - 3abc$
$\qquad =\:3 + 0 - (\text{-}1) - 3(1) \;=\;1$

The denominator simplifies to:
$\quad 1 + (a+b+c) + (ab+bc+ac) + abc$
$\qquad =\:1+0+(\text{-}1)+1 \;=\;1$

Therefore: $\:Y \;=\;\dfrac{1}{1} \;=\;1$
[/sp]

Spoiler

neat and elegent

 

[anemone]

Hello, anemone![sp]
Since $a,\,b,\,c$ are roots of the cubic: $\;\begin{Bmatrix}a+b+c &=& 0 \\ ab + bc + ac &=& \text{-}1 \\ abc &=& 1 \end{Bmatrix}$Then: $\,Y \,=\,\dfrac{(1-a)(1+b)(1+c) + (1+a)(1-b)(1+c) + (1+a)(1+b)(1-c)}{(1+a)(1+b)(1+c)} $

The numerator simplifies to:
$\quad 3 + (a+b+c) - (ab + bc + ac) - 3abc$
$\qquad =\:3 + 0 - (\text{-}1) - 3(1) \;=\;1$

The denominator simplifies to:
$\quad 1 + (a+b+c) + (ab+bc+ac) + abc$
$\qquad =\:1+0+(\text{-}1)+1 \;=\;1$

Therefore: $\:Y \;=\;\dfrac{1}{1} \;=\;1$
[/sp]

Well done, soroban! Very neatly done instead!(Yes) And thanks for participating!:)

Spoiler

$\dfrac{1-a}{1+a}=1- \dfrac{2a}{1+a}=1-\dfrac{2}{1+\dfrac{1}{a}}$
$a,\,b,\,c$ are the roots of $x^3-x-1=0$
so $\dfrac{1}{a},\,\dfrac{1}{b},\,\dfrac{1}{c}$ are the roots of $\dfrac{1}{x^3}-\dfrac{1}{x}-1=0$
or $x^3+x^2-1=0$
so $1+\dfrac{1}{a},\,1+\dfrac{1}{b},\,1+\dfrac{1}{c}$ are the roots of
$(x-1)^3+(x-1)^2-1=0$
or $x^3-3x^2+3x-1 +x^2-2x+1-1=0$
or $x^3-2x^2+x-1=0$
so $\dfrac{1}{1+\dfrac{1}{a}},\,\dfrac{1}{1+\dfrac{1}{b}},\,\dfrac{1}{1+\dfrac{1}{c}}$ are the roots of
$\dfrac{1}{x^3}-\dfrac{2}{x^2}+\dfrac{1}{x}-1=0$
or
$x^3-x^2+2x-1= 0$
so $\dfrac{1}{1+\dfrac{1}{a}}+\,\dfrac{1}{1+\dfrac{1}{b}}+\,\dfrac{1}{1+\dfrac{1}{c}}= 1$
or $\dfrac{a}{1+a}+\,\dfrac{b}{1+b}+\,\dfrac{c}{1+c}= 1$
or $\dfrac{2a}{1+a}+\,\dfrac{2b}{1+b}+\,\dfrac{2c}{1+c}= 2$
or $1- \dfrac{2a}{1+a}+1- \,\dfrac{2b}{1+b}+1-\,\dfrac{2c}{1+c}= 3-2$
or $\dfrac{1-a}{1+a}+\,\dfrac{1-b}{1+b}+\,\dfrac{1-c}{1+c}= 1$

Hi kaliprasad, your solution is great as well, thanks for participating, kali!

Another method that is quite similar to kali's method:

Spoiler

$\begin{align*}\dfrac{1-a}{1+a}+\dfrac{1-b}{1+b}+\dfrac{1-c}{1+c}&=\dfrac{1+a-2a}{1+a}+\dfrac{1+b-2b}{1+b}+\dfrac{1+c-2c}{1+c}\\&=3-2\left(\dfrac{a}{1+a}+\dfrac{b}{1+b}+\dfrac{c}{1+c}\right)\\&=3-2\left(\dfrac{1+a-1}{1+a}+\dfrac{1+b-1}{1+b}+\dfrac{1+c-1}{1+c}\right)\\&=3-6+2\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)\\&=2\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)-3\end{align*}$

We're told that $a,\,b,\,c$ are the roots of $f(x)=x^3-x-1$, hence, the function $f(x-1)=(x-1)^3-(x-1)-1=x^3-3x^2+2x-1$ has roots $a+1,\,b+1,\,c+1$ and hence the sum of the reciprocal roots of $a+1,\,b+1,\,c+1$ is 2.

$\therefore \dfrac{1-a}{1+a}+\dfrac{1-b}{1+b}+\dfrac{1-c}{1+c}=2\left(\dfrac{1}{1+a}+\dfrac{1}{1+b}+\dfrac{1}{1+c}\right)-3=2(2)-3=1$