Solving a system of equations: sum of 3 variables is 2; sum of their squares is 6; sum of their cubes is 8

[Aion]

Problem. Determine $x_1,x_2,x_3$ if $$x_1+x_2+x_3=2$$ $$x_1^2+x_2^2+x_3^2=6$$ $$x_1^3+x_2^3+x_3^3=8$$
My solution attempt.

Suppose $p(x)$ is a polynomial of the third degree with three roots $x_1,x_2,x_3$ that satisfy the three conditions above. Then by the factor theorem, we can conclude that $p(x)=q(x)(x-x_1)(x-x_2)(x-x_3)$. Since it's of the third degree, $q(x)=1$. By expanding the expression we obtain

$$p(x)=x^3+a_1x^2+a_2x+a_3=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3$$
We note that $a_1=-(x_1+x_2+x_3)=-2$. By a corollary to Vieta's theorem $x_1^2+x_2^2+x_3^2=a_1^2-2a_2$, hence $a_2=-1$. By doing some algebra manipulations we can show that the following identity holds.

$$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1x_2+x_1x_3+x_2x_3)(x_1+x_2+x_3)+3x_1x_2x_3$$
By substitution of known values into this expression, we conclude that $x_1x_2x_3=-2$, hence $a_3=2$. The polynomial equation that satisfies the three conditions are therefore

$$p(x)=x^3-2x^2-x+2=(x-2)(x-1)(x+1)=0$$
Thus the roots of the equation are $x_1=2,x_2=1,x_3=-1$ (without regard to index due to symmetry).


My questions are the following. Is there any easier way to solve this system of equations? And how would you prove that the solution set is unique?

 

[Aion]

Problem. Determine $x_1,x_2,x_3$ if $$x_1+x_2+x_3=2$$ $$x_1^2+x_2^2+x_3^2=6$$ $$x_1^3+x_2^3+x_3^3=8$$
My solution attempt.

Suppose $p(x)$ is a polynomial of the third degree with three roots $x_1,x_2,x_3$ that satisfy the three conditions above. Then by the factor theorem, we can conclude that $p(x)=q(x)(x-x_1)(x-x_2)(x-x_3)$. Since it's of the third degree, $q(x)=1$. By expanding the expression we obtain

$$p(x)=x^3+a_1x^2+a_2x+a_3=x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3$$
We note that $a_1=-(x_1+x_2+x_3)=-2$. By a corollary to Vieta's theorem $x_1^2+x_2^2+x_3^2=a_1^2-2a_2$, hence $a_2=-1$. By doing some algebra manipulations we can show that the following identity holds.

$$x_1^3+x_2^3+x_3^3=(x_1+x_2+x_3)^3-3(x_1x_2+x_1x_3+x_2x_3)(x_1+x_2+x_3)+3x_1x_2x_3$$
By substitution of known values into this expression, we conclude that $x_1x_2x_3=-2$, hence $a_3=2$. The polynomial equation that satisfies the three conditions are therefore

$$p(x)=x^3-2x^2-x+2=(x-2)(x-1)(x+1)=0$$
Thus the roots of the equation are $x_1=2,x_2=1,x_3=-1$ (without regard to index due to symmetry).


My questions are the following. Is there any easier way to solve this system of equations? And how would you prove that the solution set is unique?

I think it's a unique solution since $p(x)$ is prime-factored, and such a prime factorization is always unique.

 

[blamocur]

I cannot think of a better solution. You might shorten it slightly:

Suppose p(x)p(x)p(x) is a polynomial of the third degree with three roots x1,x2,x3x_1,x_2,x_3x1,x2,x3 that satisfy the three conditions above. Then by the factor theorem, we can conclude that p(x)=q(x)(x−x1)(x−x2)(x−x3)p(x)=q(x)(x-x_1)(x-x_2)(x-x_3)p(x)=q(x)(x−x1)(x−x2)(x−x3). Since it's of the third degree, q(x)=1q(x)=1q(x)=1.

I'd just say "consider polynomial [imath]p(x) = (x-x_1)(x-x_2)(x_x_3) in which \(x_1,x_2,x_3[/imath] satisfy the equations above.
I also agree with you proof of uniqueness.

 

[lookagain]

[imath]p(x) = (x-x_1)(x-x_2)(x_x_3) in which \(x_1,x_2,x_3[/imath]

$... \ \ p(x) = (x - x_1)(x - x_2)(x - x_3) \ \ in \ \ which \ \ \ (x_1, x_2, x_3) \ \ ...$ <--- This is my redo of yours, but I added
more spaces and the missing closing parenthesis just before the bracket with the /imath inside of it.

blamocur, your own Latex was not coming out, because you had this

\(x_1 instead of \ (x_1 with the extra needed space.

 

[blamocur]

blamocur, your own Latex was not coming out,

Sorry for the typo -- glad you got it right.