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.

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


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 $$t^3-(a+b+c)t^2+(ab+bc+ca)t-abc=0$$

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

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


$$x+y+z=3\sqrt2u=6,\\ u=\sqrt2.$$ (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 $(1,2,3)$.


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.