0
$\begingroup$

Question: If $x_1,x_2,x_3$ are such that $x_1+x_2+x_3=2$ , $x_1^2+x_2^2+x_3^2=6$ and $x_1^3+x_2^3+x_3^3=8$. Then find the value of $-(x_2-x_3)(x_3-x_1)(x_1-x_2)$.

My attempt: Tried to correlate the values of $x_1,x_2,x_3$ but no fruitful result yet. Idk why they look like some polynomial roots to me, but I can be wrong in this context. Please help

$\endgroup$
  • $\begingroup$ could it that the result is $6$? $\endgroup$ – Dr. Sonnhard Graubner Apr 14 '17 at 7:36
  • $\begingroup$ $x_1=2,x_2=1,x_3=-1$ $\endgroup$ – Navin Apr 14 '17 at 7:37
  • $\begingroup$ @Dr.SonnhardGraubner Yes. How? $\endgroup$ – The Dead Legend Apr 14 '17 at 7:38
  • $\begingroup$ eliminating $x_2,x_3$ we get for $x_1: -x_1^3+2x_1^2+x_1-2=0$ $\endgroup$ – Dr. Sonnhard Graubner Apr 14 '17 at 7:41
  • 2
    $\begingroup$ The expression is not symmetrical, I don't think you've enough information (all the known expressions are symmetrical). $\endgroup$ – Ng Chung Tak Apr 14 '17 at 7:54
3
$\begingroup$

Yes, you are right. We have $$x_1x_2+x_2x_3+x_3x_1=\frac12\Big((x_1+x_2+x_3)^2-(x_1^2+x_2^2+x_3^2)\Big)=-1.$$ $$x_1x_2x_3=\frac16\Big((x_1+x_2+x_3)^3-3(x_1^2+x_2^2+x_3^2)(x_1+x_2+x_3)+2(x_1^3+x_2^3+x_3^3)\Big)=-2.$$ Thus by Vieta's theorem, $x_1,x_2,x_3$ are roots of the polynomial $$x^3-2x^2-x+2=(x-2)(x-1)(x+1).$$


In fact, for $n$ complex numbers $x_1,\ldots,x_n$, if given $\sum_{i=1}^{n}x_i^k$ for $k=1,\ldots,n$, then we can construct a polynomial of degree $n$ such that $x_1,\ldots,x_n$ are precisely the roots of the polynomial (and thus uniquely determined). In the general case, we will use Newton's identities.

$\endgroup$

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.