# Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself

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?

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

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

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 $$\color{red}{\frac ab+\frac bc+\frac ca=Q=-6}$$

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

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)

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

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

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.

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 $m=-1?$

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

• please use latex for answering. – Ali Jul 22 '15 at 18:17
• $\dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a}$ $\neq$ $\dfrac{a^2bc+b^2ac+c^2ab}{abc}$ As you said in the beginning. – Nikhil Jul 22 '15 at 18:21
• 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. – Ali Jul 22 '15 at 18:23