# What will be $-(x_2-x_3)(x_3-x_1)(x_1-x_2)$?

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

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

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.