How to find $x^4+y^4+z^4$ from equation? Please help me.
There are equations: $x+y+z=3, x^2+y^2+z^2=5$ and $x^3+y^3+z^3=7$. The question:
what is the result of $x^4+y^4+z^4$?
Ive tried to merge the equation and result in desperado. :(
Please explain with simple math as I'm only a junior high school student. Thx a lot
 A: use this 
since
$$x^2+y^2+z^2=(x+y+z)^2-2xy-2yz-2xz\Longrightarrow xy+yz+xz=2$$
$$x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-xz)=(x+y+z)^3-3(xy+yz+xz)(x+y+z)$$
so
$$7-3xyz=27-18\Longrightarrow xyz=-\dfrac{2}{3}$$
use
$$x^4+y^4+z^4=(x+y+z)(x^3+y^3+z^3)-(xy+yz+xz)(x^2+y^2+z^2)+xyz(x+y+z)$$
A: solution 2: since 
$$x^4+y^4+z^4=(x^2+y^2+z^2)^2-2x^2y^2-2y^2z^2-2x^2z^2$$
and
$$(xy+yz+xz)^2=x^2y^2+x^2z^2+y^2z^2+2xyz(x+y+z)$$
since
$$xy+yz+xz=2,xyz=-\dfrac{2}{3}$$
so $$x^2y^2+y^2z^2+x^2z^2=8$$
so
$$x^4+y^4+z^4=5^2-2\cdot 8=9$$
A: We have in general if
$$
s_1:=x+y+z\textrm{, }s_2:=x^2+y^2+z^2\textrm{, }s_3:=x^3+y^3+z^3
$$
and
$$
s_4:=x^4+y^4+z^4
$$
and if 
$$
\sigma_1=x+y+z\textrm{, }\sigma_2=xy+yz+zx\textrm{, }\sigma_3=xyz\tag 1
$$
Then
$$
s_1=\sigma_1\textrm{, }s_2=\sigma_1^2-2\sigma_2\textrm{, }s_3=\sigma_1^3-3\sigma_1\sigma_2+3\sigma_3\textrm{, }s_4=\sigma_1^4-4\sigma_1^2\sigma_2+2\sigma_2^2+4\sigma_1 \sigma_3\tag 2
$$
Hence if $s_1=3,s_2=5,s_3=7$, then
$$
\sigma_1=s_1=3\textrm{, }\sigma_2=2\textrm{, }\sigma_3=-2/3
$$
Hence
$$
s_4=x^4+y^4+z^4=9
$$
For to find $x^5+y^5+z^5$, we use
$$
(x+y+z)^5=x^5+y^5+z^5+20xyz(x^2+y^2+z^2)+30xyz(xy+yz+zx)+5 S,
$$
where
$$
S=\sum_{cyc}\left(x^4y+xy^4+2x^3y^2+2x^2y^3\right)
=\sum_{cyc}xy\left(x^3+y^3\right)+2x^2y^2z^2\sum_{cyc}\frac{x+y}{z^2}=
$$
$$
=\sum_{cyc}xy\left(x^3+y^3+z^3-z^3\right)+2\sigma_3^2\sum_{cyc}\frac{s_1-z}{z^2}=
$$
$$
=s_3\sum_{cyc}xy-\sigma_3s_2+2\sigma_3^2s_1\sum_{cyc}x^{-2}-2\sigma_3^2\sum_{cyc}x^{-1}=
$$
$$
=s_3\sigma_2-\sigma_3s_2+2\sigma_3^2s_1\frac{-\sigma_2^2+2\sigma_1\sigma_2}{\sigma_3^2}-2\sigma_3^2\frac{\sigma_2}{\sigma_3}.
$$
Since
$$
\sum_{cyc}\frac{1}{x^2}=\frac{2\sigma_1\sigma_2-2\sigma_2^2}{\sigma_3^2}\textrm{, }\sum_{cyc}\frac{1}{x}=\frac{\sigma_2}{\sigma_3}.
$$
Hence finaly in general
$$
s_5=x^5+y^5+z^5=\sigma_1^5-5\sigma_1^3\sigma_2+25\sigma_1\sigma_2^2-5\sigma_2\sigma_3-5\sigma_1^2(4\sigma_2+3\sigma_3).\tag 3
$$
and as application
$$
x^5+y^5+z^5=29/3.
$$
