A simple problem with a simple and a nonsimple solution I have this seemingly simple problem to solve. Its statement is straightforward:
"If $x+y+z=3,x^2+y^2+z^2=5, x^3+y^3+z^3=7$ show that $x^4+y^4+z^4=9,x^5+y^5+z^5\neq 11$". 
There is a highschool approach, that is I expanded $(x+y+z)^2, (x+y+z)^3$ and managed to get $$xyz=-2/3, xy+yz+zx=2$$ etc, which I guess I could employ to expand $(x+y+z)^4, (x+y+z)^5$ and get some results (I guess).
The thing is I don't think this is the proper approach. Is there a better way than this tedious, long calculation (if it is  a solution that is)
Say we could divide the polynomials
$$f=x^4+y^4+z^4,g=x^3+y^3+z^3, h=x^2+y^2+z^2,k=x+y+z$$
Wouldn't this give us what we want by using the remainder polynomial?
Thanks in advance for your time. 

$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2(x^2y^2+x^2z^2+y^2z^2)=$$
$$=25-2((xy+xz+yz)^2-2xyz(x+y+z))=$$
$$=25-2\left(4-2\left(-\frac{2}{3}\right)3\right)=9.$$
$$x^5+y^5+z^5=(x^2+y^2+z^2)(x^3+y^3+z^3)-\sum_{sym}x^3y^2=$$
$$=35-(x^2y^2+x^2z^2+y^2z^2)(x+y+z)+xyz(xy+xz+yz)=$$
$$=35-8\cdot3-\frac{2}{3}\cdot2=\frac{29}{3}$$
