6
$\begingroup$

Suppose that we have a equation of third degree as follows:

$$ x^3-3x+1=0 $$

Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?

$$ \frac {a}{b} + \frac {b}{c} + \frac {c}{a} $$

$\endgroup$
  • $\begingroup$ the answer should be $3$ $\endgroup$ – Dr. Sonnhard Graubner Jul 22 '15 at 17:42
  • 1
    $\begingroup$ Main problem is that the expression is not symmetric ... $\endgroup$ – Hagen von Eitzen Jul 22 '15 at 17:42
  • 1
    $\begingroup$ Put $x=2\cos y$ $\endgroup$ – lab bhattacharjee Jul 22 '15 at 17:44
5
$\begingroup$

By Vieta's formulas, we have $$a+b+c=-\frac{0}{1}=0\tag1$$ $$ab+bc+ca=\frac{-3}{1}=-3\tag2$$ $$abc=-\frac{1}{1}=-1\tag3$$ From $(1)(2)(3)$, we have $$P=a^2b+ab^2+b^2c+bc^2+c^2a+ca^2=(a+b+c)(ab+bc+ca)-3abc=3\tag4$$

Now, set $$Q=\frac ab+\frac bc+\frac ca,\ \ \ R=\frac ba+\frac cb+\frac ac.$$ Then, we have $$Q+R=\frac ab+\frac bc+\frac ca+\frac ba+\frac cb+\frac ac=\frac{P}{abc}=-3\tag5$$ and $$\begin{align}QR&=\left(\frac ab+\frac bc+\frac ca\right)\left(\frac ba+\frac cb+\frac ac\right)\\&=3+\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}+\frac{c^2}{ab}+\frac{a^2}{bc}+\frac{b^2}{ca}\\&=3-\frac{1}{a^3}-\frac{1}{b^3}-\frac{1}{c^3}+\frac{a^3+b^3+c^3}{abc}\\&=3-\frac{1}{3a-1}-\frac{1}{3b-1}-\frac{1}{3c-1}+\frac{3abc+(a+b+c)((a+b+c)^2-3(ab+bc+ca))}{abc}\\&\small=3+\frac{-9(ab+bc+ca)+6(a+b+c)-3}{27abc-9(ab+bc+ca)+3(a+b+c)-1}+\frac{3abc+(a+b+c)((a+b+c)^2-3(ab+bc+ca))}{abc}\\&=-18\tag6\end{align}$$

Also, since it is easy to see $$a\lt 0\lt b\lt c,$$ we have $$\frac ab\lt 0,\frac bc\lt 1,\frac ca\lt 0\Rightarrow Q\lt 1\tag 7$$

So, as a result, from $(5)(6)(7)$ we have $$\color{red}{\frac ab+\frac bc+\frac ca=Q=-6}$$

$\endgroup$
  • $\begingroup$ $a=2\cos160^\circ,b=2\cos80^\circ,c=2\cos40^\circ$ and see here. $\endgroup$ – mathlove Jul 22 '15 at 19:13
6
$\begingroup$

We have $$ \frac {a}{b} + \frac {b}{c} + \frac {c}{a} = \frac{a^2c+b^2a+c^2b}{abc} $$ Since from Vieta's relations we know $$ a+b+c =0,\quad ab+bc+ca =-3,\quad abc =-1, $$ our goal is to calculate $$ s = a^2c+b^2a+c^2b. $$ Let's introduce $$ p = ac^2+ba^2+cb^2. $$ Than we have $$ 0 = (ab+bc+ac)(a+b+c) = p+s+3abc $$ and $p+s = 3$.

Now let's multiply $$ s\cdot p = a^3b^3 + a^3c^3 +b^3c^3 + 3(abc)^2+ abc(a^3+b^3+c^3) $$ Since $a,b,c$ are the roots of polynomial the last equation can be rewritten as $$ sp = (3a-1)(3b-1)+(3a-1)(3c-1)+(3b-1)(3c-1) + 3(abc)^2 + abc(3(a+b+c)-3)= $$ $$ =9(ab+ac+bc)-6(a+b+c)+3 + 3(abc)^2 + abc(3(a+b+c)-3) =-27+3+3+3=-18. $$

So, $s+p=3$ and $sp=-18$. From here one can deduce that $s= 6$. (see @mathlove answer)

$\endgroup$
  • $\begingroup$ Nice work. Your way is simpler than mine, but $sp\not=0$, I think. Also, no two possibilities because of $a\lt b\lt c$. $\endgroup$ – mathlove Jul 22 '15 at 19:36
  • $\begingroup$ @mathlove yes, you are right, $sp \neq 0$, I have made a mistake $\endgroup$ – Virtuoz Jul 22 '15 at 19:45
2
$\begingroup$

If you multiply out the expression $(x-a)(x-b)(x-c)$ and compare the coefficients to the expression after you get a common denominator, all will become clear

$\endgroup$
1
$\begingroup$

We know that every symmetric function of the roots $a,b,c$ can be evaluated in terms of the elementary symmetric functions: $$ e_1=a+b+c=0,\quad e_2=ab+ac+bc=-3,\quad e_3=abc=-1$$ or the power sums: $$ p_1=e_1=0,\quad p_2=a^2+b^2+c^2 = 6,\quad p_3=a^3+b^3+c^3=3e_1-3=-3.$$ Now: $$ g(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=-(a^2 c+b^2 a+c^2 b)$$ is not a symmetric function of $a,b,c$, and neither it is: $$ h(a,b,c)=\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=-(a^2 b+b^2 c+c^2 a),$$ but both $g+h$ and $g\cdot h$ are. So the strategy is just to find $g+h$ and $g\cdot h$ in terms of $e_1,e_2,e_3$, then solve a quadratic equation to find $\{g,h\}$ and recognize $g$ from the constraint $a<b<c$.

We have: $$\begin{eqnarray*} g+h &=& -(a^2(b+c)+b^2(a+c)+c^2(a+b))\\ &=& (a^3+b^3+c^3)-(a^2+b^2+c^2)(a+b+c)\\&=&p_3-p_2 p_1=-3,\end{eqnarray*}$$

$$\begin{eqnarray*} g\cdot h &=& e_3 p_3 + 3a^2b^2c^2 + e_3^3\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\\&=&6-\left(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}-3\right)\\&=&9-3\left(\frac{e_2^2}{e_3^2}-2\frac{e_1}{e_3}\right)=-18,\end{eqnarray*}$$ hence $g,h$ are the roots of $z^2+3z-18$, and $\{g,h\}=\{-6,3\}$. Since $e_3<0$, we have $a<0<b<c$, from which: $$ -g = a^2 c+b^2 a+ c^2 b = (b+c)^2 c-b^2(b+c)+c^2 b = c^3-b^3+3bc^2 > 0 $$ and $\color{red}{g=-6}$ follows.

$\endgroup$
1
$\begingroup$

As I've suggested in the comment yesterday,

let $x=2m\cos y\implies(2m\cos y)^3-(2m\cos y)+1=0\ \ \ \ (1)$

As $\cos3y=4\cos^3y-3\cos y,$

$2m^3(\cos3y+3\cos y)-(2m\cos y)+1=0$

$\iff2m^3\cos3y+2m\cos y(m^2-1)+1=0\ \ \ \ (2)$

WLOG choose $m^2-1=0\iff m=\pm1$

Let $m=1$

$(1)$ reduces to $8\cos^3y-6\cos y+1=0 \ \ \ \ (3)$

and $(2)\implies\cos3y=-\dfrac12\implies3y=360^\circ n\pm120^\circ$ where $n$ is any integer

$\implies y=120^\circ n+40^\circ$ where $n\equiv-1,0,1\pmod3$

So, the roots of $(3)$ are

$\cos(-80^\circ)=\cos80^\circ$ $\cos40^\circ,\cos160^\circ=\cos(180^\circ-20^\circ)=-\cos20^\circ<0$

Clearly, $\cos40^\circ>\cos80^\circ>0>-\cos20^\circ$

$\implies c=2\cos40^\circ, b=2\cos80^\circ, a=2\cos160^\circ$

$\implies\dfrac ab=\dfrac{2\cos160^\circ}{2\cos80^\circ}=\dfrac{2\cos^280^\circ-1}{\cos80^\circ}=2\cos80^\circ-\dfrac1{\cos80^\circ}$

$\implies\sum_{\text{cyc}}\dfrac ab=2\sum_{\text{cyc}}\cos80^\circ-\sum_{\text{cyc}}\dfrac1{\cos80^\circ}$

Using Vieta's formula on $(3),\sum_{\text{cyc}}\cos80^\circ=0,$

$\cos40^\circ\cos80^\circ+\cos40^\circ\cos160^\circ+\cos80^\circ\cos160^\circ=\dfrac{-6}8$

and $\cos40^\circ\cos80^\circ\cos160^\circ=-\dfrac18$

and $\sum_{\text{cyc}}\dfrac1{\cos80^\circ}=\dfrac{\cos40^\circ\cos80^\circ+\cos40^\circ\cos160^\circ+\cos80^\circ\cos160^\circ}{\cos40^\circ\cos80^\circ\cos160^\circ}=\cdots=6$

Won't you try with $m=-1?$

$\endgroup$
-4
$\begingroup$

Required, $= a/b + b/c + c/a$

By Cross multiplication,

$= (a^2bc+b^2ac+c^2ab)/(abc)$

$= abc (a+b+c) /(abc)$

$ = (a+b+c) ..........1$

Well known

Property Relation:-

{If α1, α2,α3 ... αn are the roots of the equation

$f(x)= a_0x_n +a_1x_{n-1} +a_2x_{n-2} +...+a_{n-1}x + a_n =0$, then

$f(x)= a_0 (x-α_1)(x-α_2)(x-α_3)... (x-α_n)$

Equating both the RHS terms we get,

$a_0x_n +a_1x_{n-1} +a_2x_{n-2} +...+a_{n-1}x + a_n = a_0(x-α_1)(x-α_2)(x-α_3)... (x-α_n)$

Comparing coefficients of $x_{n-1}$ on both sides, we get

  $S1 = α_1 + α_2+α_3 +... + α_n = ∑α_i  = -a_1/ a_0$ 

or, S1= - coeff. of $x_n-1$/coeff. of $x_n$

Comparing coefficients of xn-2 on both sides, we get

  $S2 = α_1 α_2+ α_1α_3 +...  = ∑α_i α_j  = (-1)^2a_2/ a_0$ 
                                         $i≠ j$

or, S2= (-1)2 coeff. of $x_{n-2}$/coeff. of $x_n$

Comparing coefficients of xn-3 on both sides, we get

  $S3 = α_1 α_2α_3+ α_2α_3α_4 +...  = ∑α_i α_j α_k  = (-1)^3a_3/ a_0$ 

If $a$ $b$ $c$ are the roots of $x^3-3x+1=0$

then

$abc = 1$

$a+b+c = 0$

$ab + bc + cd = -1$

Substituting in eqn 1, we get Ans: 0

$\endgroup$
  • 1
    $\begingroup$ please use latex for answering. $\endgroup$ – Ali Jul 22 '15 at 18:17
  • $\begingroup$ $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ $\neq$ $\dfrac{a^2bc+b^2ac+c^2ab}{abc}$ As you said in the beginning. $\endgroup$ – Nikhil Jul 22 '15 at 18:21
  • $\begingroup$ It seems that the answer is wrong. we have $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=\frac{a^2c+b^2a+c^2b}{abc}$$ but Shubham Kulkarni wrote noncorect. $\endgroup$ – Ali Jul 22 '15 at 18:23

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.