# Finding trinity of complex numbers

I want to find all $x_1,x_2,x_3$ that satisfy these three equalities: $$x_1+x_2+x_3=6$$ $$x_1^2+x_2^2+x_3^2=14$$ $$x_1^3+x_2^3+x_3^3=36$$

So i don't know whether i should solve it using the same techniques i do with normal systems of equations or is there some tricks behind it?

So since I found this problem while doing algebraic problems I think there should be an algebraic solution to it. But I am really lost in this case, I don't knwow how to begin.

Any help would be appreciated.

• Mh, by inspection... then permutation. – Yves Daoust Feb 10 '17 at 10:25

HINT:

Use $$(a+b+c)^2=(a^2+b^2+c^2)+2(ab+bc+ca)$$

and $$a^3+b^3+c^3-3abc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}$$

to find $ab+bc+ca, abc$

Then $a,b,c$ are the roots of $$t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$

• But what is $t$? – MathIsTheWayOfLife Feb 10 '17 at 10:17
• @MathIsTheWayOfLife, The variable of the cubic equation, whose values are required – lab bhattacharjee Feb 10 '17 at 10:18
• oh right, so it's solvable with what i learned about polynomial in algebra, thank you – MathIsTheWayOfLife Feb 10 '17 at 10:18
• How many solutions do you get ? – Yves Daoust Feb 11 '17 at 13:38
• @YvesDaoust, math.stackexchange.com/questions/25822/… – lab bhattacharjee Feb 11 '17 at 14:01

The hard way:

The first equation is that of a plane with normal $(1,1,1)$ and this direction is that of an axis of symmetry of the problem.

We will rotate it to bring it to a $u$ axis, using the similarity transform

$$\begin{cases}x=\sqrt2 u-\sqrt3 v+w,\\y=\sqrt2u+\sqrt3v+w,\\z=\sqrt2u\ \ \ \ \ \ \ \ \ \ -2w.\end{cases}$$

Then

$$x+y+z=3\sqrt2u=6,\\ u=\sqrt2.$$ (A plane perpendicular to $u$.) $$x^2+y^2+z^2=6(u^2+v^2+w^2)=14,\\v^2+w^2=\frac13.$$ (A cylindre of axis $u$.) $$x^3+y^2+z^3=18\sqrt2uv^2+18\sqrt2uw^2+6\sqrt2u^3+18wv^2-6w^3\\ =18\sqrt2uv^2+18\sqrt2uw^2+18\sqrt2uu^2-12\sqrt2u^3+18wv^2-6w^3\\ =42\sqrt2u-12\sqrt2u^3+18wv^2-6w^3=36,\\ 3wv^2=w^3.$$ (Three planes intersecting on $u$.)

This gives the six solutions

$$v=\pm\frac1{\sqrt3},w=0\lor v=\pm\frac1{2\sqrt3},w=\pm\frac12,$$ i.e. $$\begin{cases}x=2\mp1,\\ y=2\pm1,\\ z=2,\end{cases}$$ or $$\begin{cases}x=2\mp\dfrac12\pm\dfrac12,\\ y=2\pm\dfrac12\pm\dfrac12,\\ z=2\ \ \ \ \ \ \ \ \ \mp1.\end{cases}$$

As expected, the six permutations of $(1,2,3)$.