I should find $A,B$ and $C$. I know answers but can't figure out how to solve it. Anyone?
We are to find value of $x^4+y^4+z^4$ when $x, y$ and $z$ are real numbers which satisfy the following three equalities: $$x+y+z=3$$ $$x^2+y^2+z^2=9$$ $$xyz=-2$$
Firstly, it follows from the first two equalities that $$xy+yz+zx=A$$
Next, using $$(x^2+y^2+z^2)^2=x^4+y^4+z^4+B[(xy)^2+(yz)^2+(zx)^2],$$ we have $x^4+y^4+z^4=C$.
Use Newton-Girard relations between elementary symmetric finctions and sums of powers:
Let $p_k=x^k+y^k+z^k$, $\,s_1=x+y+z$, $\,s_2=xy+yz+zx$, $\,s_3=xyz$. We have: \begin{align*} p_1&=s_1& p_2&=s_1p_1-2s_2=s_1^2-2s_2\\ p_3&=s_1p_2-s_2p_1+3s_3&p_4&=s_1p_3-s_2p_2+s_3p_1 \end{align*} We deduce that $s_2=\frac12(s_1^2-p_2)=0$, whence $\,p_3= 21$ and finally $\,p_4=63-6=57$.