Question

Solve the following equations :
$x+y+z=12,x^2+y^2+z^2=50,x^3+y^3+z^3=216.$

Answer 1

From first two equations, we get
$xy+yz+zx=\frac{1}{2}[\left({x+y+z}^2\right)-(x^2+y^2+z^2)]=\frac{1}{2}[144-50]=47$
Also $x^3+y^3-3xyz=(x+y+z)(x^2+y^2+z^2-yz-zx-xy)$
or $216-3xyz=12(50-47)$
or $xyz=60$.
Hence the given equations reduce to
$x+y+z=12,yz+zx+xy=47,xyz=60$.
Now proceed as in problem 23.
Solution sets in this case are:
$(3,4,5),(3,5,4),(4,3,5),(4,5,3),(5,3,4)$ and $(5,4,3)$
Alt. Here.$S_1=12$. Also $x^2+y^2+z^2=50$ gives
${(x+y+z)}^2-2(xy+yz+zx)=50$
or $144-2S_2=50$ $\therefore 2S_2=94$ or $S_2=47$
Again $x^3+y^3+z^3=216$.
But $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-\sum xy)$
or $216-3S_3=12(50-47)=36$
or $216-36=3S_3$ $\therefore S_3=60$
$\therefore S_1=12,S_2=47,S_3=60$
$\therefore x,y,z$ arethe roots of cubic in$t$:
$t^3-S_1{t}^2+S_{2}t-S_3=0$
or $t^3-12t^2+47t-60=0$
By trial $t=3$ satisfies it.
$\therefore (t-3)(t^2-9t+20)=4$
or $(t-3)(t-4)(t-5)=0$
$\therefore t=3,4,5$.
Hence the solution sets as above are
$\begin{array}{c}(3,4,5)\\ (3,5,4)\end{array}$ or $\begin{array}{c}(4,3,5)\\ (4,5,3)\end{array}$ or $\begin{array}{c}(5,3,4)\\ (5,4,3)\end{array}$