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 $a &lt; b &lt; c$ holds. How can we find the answer of the following expression, without solving the original equation?

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