<s>[INST] 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?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
 [/INST] 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
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