7
$\begingroup$

Find the value of

$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3$
where $\alpha,\beta,\gamma$ are roots of the equation $x^3+2x^2+3x+3=0$.

I tried to use the formula which is wrong

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

And then I broke the terms to get

$(\frac{13}{5})^3-\frac{36}{5}$

but this is $\ne 44$, which should be the answer.

I know where I was wrong, the formula is $(a+b+c)^3=a^3+b^3+c^3-3(a+b)(b+c)(c+a)$

$\endgroup$
2
  • 2
    $\begingroup$ I don't believe your formula $$(\frac{\alpha}{\alpha +1})^3+(\frac{\beta}{\beta +1})^3+(\frac{\gamma}{\gamma +1})^3=((\frac{\alpha}{\alpha +1})+(\frac{\beta}{\beta +1})+(\frac{\gamma}{\gamma +1}))^3-3((\frac{\alpha}{\alpha +1})(\frac{\beta}{\beta +1})(\frac{\gamma}{\gamma +1}))$$ it is missing a lot of terms of the RHS like $\dfrac{\alpha^2\beta}{(1+\alpha)^2(1+\beta)}$ and the coefficient of $\dfrac{\alpha\beta\gamma}{(1+\alpha)(1+\beta)(1+\gamma)}$ isn't correct $\endgroup$ Commented Nov 4, 2018 at 5:15
  • $\begingroup$ @user10354138 Please check the edit where I have written the correct formula .Sorry for that. $\endgroup$
    – Saradamani
    Commented Nov 4, 2018 at 7:02

4 Answers 4

6
$\begingroup$

Substituting $x=\frac y{1-y}$, which is the inverse of $y=\frac x{x+1}$, yields $$y^3-5y^2+6y-3=0$$ and $a,b,c=\frac\alpha{1+\alpha},\frac\beta{1+\beta},\frac\gamma{1+\gamma}$ are its roots. Then the desired expression is $$a^3+b^3+c^3=(a+b+c)^3-3(a+b+c)(ab+bc+ca)+3abc$$ and we can extract the values of the symmetric polynomials from the coefficients: $a+b+c=5,ab+bc+ca=6,abc=3$ and $$a^3+b^3+c^3=5^3-3\cdot5\cdot6+3\cdot3=44$$

$\endgroup$
1
  • $\begingroup$ (+1) I did not see your answer until I posted mine. I have deleted mine, which was very close to yours. $\endgroup$
    – robjohn
    Commented Nov 4, 2018 at 7:02
3
$\begingroup$

For simplicity, let me use $\alpha=a, \beta=b,\gamma=c$.

We can find: $$\begin{cases}a+b+c=-2 \\ ab+bc+ca=3\\ abc=-3\end{cases} \Rightarrow \\ \begin{cases}\color{red}{a^2+b^2+c^2}=4-2(ab+bc+ca)=\color{red}{-2} \\ \color{blue}{a^2b^2+b^2c^2+c^2a^2}=9-2abc(a+b+c)=\color{blue}{-3}\end{cases}$$ We can express: $$x^3+2x^2+3x+3=0 \Rightarrow (x+1)^3=x^3+3x^2+3x+1=x^2-2$$ Hence: $$\left(\frac{a}{a +1}\right)^3+\left(\frac{b}{b +1}\right)^3+\left(\frac{c}{c +1}\right)^3=\\ \frac{a^3}{a^2-2}+\frac{b^3}{b^2-2}+\frac{c^3}{c^2-2}=\\ a+\frac{2a}{a^2-2}+b+\frac{2b}{b^2-2}+c+\frac{2c}{c^2-2}=\\ -2+2\cdot \frac{a(b^2-2)(c^2-2)+b(a^2-2)(c^2-2)+c(a^2-2)(b^2-2)}{(a^2-2)(b^2-2)(c^2-2)}=\\ -2+2\cdot \frac{abc(ab+bc+ca)-2(a^2b+ab^2)-2(a^2c+ac^2)-2(b^2c+bc^2)+4(a+b+c)}{a^2b^2c^2-2(\color{blue}{a^2b^2+b^2c^2+c^2a^2})+4(\color{red}{a^2+b^2+c^2})-8}=\\ -2+2\cdot \frac{-9-2ab(-2-c)-2ac(-2-b)-2bc(-2-a)-8}{9+6-8-8}=\\ -2+2\cdot \frac{-9+4(ab+bc+ca)+6abc-8}{-1}=\\ -2+2\cdot \frac{-9+12-18-8}{-1}=44.\\ $$

$\endgroup$
3
$\begingroup$

I think your way is not really practical. Here it is a shortcut: $$ \begin{array}{|c|c|}\hline \text{Roots}&\text{Polynomial}\\ \hline \alpha,\beta,\gamma & x^3+2x^2+3x+3\\ \hline \frac{1}{\alpha},\frac{1}{\beta},\frac{1}{\gamma} & x^3+x^2+\frac{2}{3}x+\frac{1}{3}\\ \hline 1+\frac{1}{\alpha},1+\frac{1}{\beta},1+\frac{1}{\gamma} & x^3-2x^2+\frac{5}{3}x-\frac{1}{3}\\ \hline \frac{\alpha}{\alpha+1},\frac{\beta}{\beta+1},\frac{\gamma}{\gamma+1}&x^3-5x^2+6x-3 \\ \hline \end{array} $$ Now by setting $A=\frac{\alpha}{\alpha+1},B=\frac{\beta}{\beta+1},C=\frac{\gamma}{\gamma+1}$ we have $A+B+C=5$ and $AB+AC+BC=6$ by Vieta's formulas, hence $A^2+B^2+C^2=25-12=13$. Additionally for any $z\in\{A,B,C\}$ we have $z^3=5z^2-6z+3$, hence

$$ A^3+B^3+C^3 = 5\cdot 13-6\cdot 5+3\cdot 3 = 65-30+9 = \color{red}{44}.$$

An equivalent way is to compute the trace of $\left(M(M+I)^{-1}\right)^3=(I+M^{-1})^{-3}$,
where $M$ is the companion matrix of $x^3+2x^2+3x+3$.

$\endgroup$
1
$\begingroup$

$$\alpha+\beta+\gamma=-2,$$ $$\alpha\beta+\alpha\gamma+\beta\gamma=3$$ and $$\alpha\beta\gamma=-3.$$ Thus, $$\sum_{cyc}\frac{\alpha}{\alpha+1}=\frac{\sum\limits_{cyc}\alpha(\beta+1)(\gamma+1)}{\prod\limits_{cyc}(\alpha+1)}=\frac{\sum\limits_{cyc}(\alpha\beta\gamma+2\alpha\beta+\alpha)}{\alpha\beta\gamma+\alpha\beta+\alpha\gamma+\beta\gamma}=\frac{-9+6-2}{-3+3-2+1}=5,$$ $$\sum_{cyc}\frac{\alpha\beta}{(\alpha+1)(\beta+1)}=\frac{\sum\limits_{cyc}\alpha\beta(\gamma+1)}{\prod\limits_{cyc}(\alpha+1)}=\frac{-9+3}{-1}=6$$ and $$\prod_{cyc}\frac{\alpha}{\alpha+1}=\frac{\alpha\beta\gamma}{\prod\limits_{cyc}(\alpha+1)}=\frac{-3}{-1}=3.$$ Id est, $$\sum_{cyc}\left(\frac{\alpha}{\alpha+1}\right)^3=\left(\sum_{cyc}\frac{\alpha}{\alpha+1}\right)^3-3\sum_{cyc}\frac{\alpha}{\alpha+1}\sum_{cyc}\frac{\alpha\beta}{(\alpha+1)(\beta+1)}+3\prod_{cyc}\frac{\alpha}{\alpha+1}=$$ $$=5^3-3\cdot5\cdot6+3\cdot3=44.$$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .