3

If $x_1, x_2, x_3, x_4$ are the roots of $x^4-x^3+2x^2+3x+1$ then find $$\frac{\sum x_i^3+\sum{x_i^3x_j^3x_k^3}}{\sum{x_i^3x_j^3}-x_1^3x_2^3x_3^3x_4^3}$$

My attempt: I have tried many approaches but this attempt seems to have brought me nearest to a solution. First,$\sum{x_i^2}$ and $\sum{x_i^3}$ can be easily calculated since $\sum{x_i}, \sum{x_ix_j}, \sum{x_ix_jx_k}$ and $x_1x_2x_3x_4$ are already known. Then, $\sum{x_i^n}$ (n>3) can be calculated using the given biquadratic equation.
for example: $\sum{x_i^4}$ can be calculated by the equation: $\sum{x_i^4}-\sum{x_i^3}+2\sum{x_i^2}+3\sum{x_i}+1=0$
Similarly, $\sum{x_i^5}$, $\sum{x_i^6}$,……..$\sum{x_i^n}$ can be calculated.
Now, using the Newton-Girard identity,
$\sum{(x_i^3)^4}$ -$\sum{(x_i^3)^3}$.$\sum{(x_i^3)}$ +$\sum{(x_i^3)^2}$.$\sum{x_i^3x_j^3}$ -$\sum{(x_i^3)}$.$\sum{x_i^3x_j^3x_k^3}$ +$x_1x_2x_3x_4$=0
I tried to find the relation between numerator and denominator terms.($\sum x_i^3, \sum{x_i^3x_j^3x_k^3}, \sum{x_i^3x_j^3}$ and $x_1^3x_2^3x_3^3x_4^3$).
But nothing seems to work. Also, this attempt was way too lengthy. I am stuck up on this question for very long. First, I was looking for an elegant solution but now any solution will do. Please suggest a direction for solving this.

Monster196883
  • 97

2 Answers2

3

Hint: Let $P(x)=x^4-x^3+2x^2+3x+1$, and let $\omega\ne1$ be a third root of unity. Then consider $Q(x^3)=P(x)P(\omega x)P(\omega^2 x)$. Convince yourself that $Q(x)$ has roots $x_i^3$ and using Vieta you can read off the terms you need from the coefficients of $Q$

Working that out, $Q(x)=x^4+14x^3+50x^2+6x+1$, so we need $\dfrac{-14-6}{50-1}=-\frac{20}{49}$.

Macavity
  • 46,381
  • Why Q(x) has roots $x_i^3$ ? I do not understand. – Monster196883 Dec 12 '19 at 14:52
  • 1
    If $a$ is a root of $P(x)$, then $a, a\omega, a\omega^2$ are roots of $P(x)P(\omega x) P(\omega^2 x)$. Also note that this polynomial will have only powers of $x$ divisible by $3$, hence it is effectively a polynomial in $x^3$, so written as a polynomial in $x$, it has $a^3$ as a root. Think on it and let me know if any particular step needs more elaboration. – Macavity Dec 12 '19 at 15:00
  • Then, If F(x) is any polynomial of m degree and has roots $x_1$, $x_2$..... $x_m$, and Q(x) is a polynomial with roots $x_1^n$, $x_2^n$..... $x_m^n$, then Q($x^n$)=F(x).F($\omega$x)……….F($\omega^(n-1)$) ,where $\omega$ is nth root of unity. What is the proof of this? – Monster196883 Dec 12 '19 at 16:56
  • Similar argument works, and $Q$ is the unique polynomial of degree $m$ upto a constant multiple. The calculation gets tedious pretty fast, so you may be equally well off by using Vieta on the original polynomial and elementary symmetric expressions. I would use this approach for transforming polynomials for squares and cubes at best. – Macavity Dec 12 '19 at 17:06
0

Both the numerator $p$ and the denominator $q$ are symmetric functions of the four zeros $x_j$. Therefore they are both expressible rationally in terms of the known symmetric functions $\sigma_1$, $\sigma_2$, $\sigma_3$,$\sigma_4$ of the $x_j$.

Solving the equation numerically and computing $p$, $q$ using the approximative values I found $p\approx-20$, $q\approx49$.

Christian Blatter
  • 226,825
  • How did you express $p$ and $q$ in terms of symmetric functions? – Monster196883 Dec 15 '19 at 08:41
  • I'm not an expert in these matters. When forced to really do it I would go along the (inductive) proof of the theorem about symmetric polynomials. For the above answer I used the actual numerical roots $x_j$ found by Mathematica, and used Mathematica to compute $p$ and $q$ numerically. It then gave the above values correct to ten decimal places. – Christian Blatter Dec 15 '19 at 09:30