How would one solve the following system with Newton's sums and Vieta's relations?: $$x+y+z=14$$ $$x^2+y^2+z^2=14$$ $$x^3+y^3+z^3=34$$
I have taken an algebra lesson in the awesome math academy, but over time I've forgotten aspects of Newtons sums. If you could provide a detailed explanation of this, I'm sure I would be able to strengthen my understanding.
$\bf{My\; Solution::}$ Let $x,y,z$ be the roots of an Cubic equation in terms of variable $t$
So $$(t-x)\cdot (t-y)\cdot (t-z) = 0\Rightarrow t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$$
Now Given $$x+y+z=4\;\;,x^2+y^2+z^2=14\;\;, x^3+y^3+z^3=34$$
So $$(x+y+z)^2 = (x^2+y^2+z^2)+2(xy+yz+zx)\Rightarrow (xy+yz+zx) = 1$$
And $$x^3+y^3+z^3-3xyz = (x^2+y^2+z^2-xy-yz-zx)\cdot (x+y+z)$$
$$\displaystyle 34-3xyz=(14-1)\cdot 4=52\Rightarrow \displaystyle xyz = -6$$
So Put these values into equation in terms of $t\;,$ We get
$$\displaystyle t^3-4t^2+t+6=0\Rightarrow (t+1)\cdot (t-2)\cdot (t-3) = 0$$
So We get $$(x,y,z)\in \left\{-1,2,3\right\}$$. Means all Arrangement of $-1,2,3$
