System of 3 equations problem 
If
  $$x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1$$
  Then how much is
  $$x^4+y^4+z^4$$

So what I'm pretty sure about, this shouldn't be solved for $x$,$y$, and $z$, but I should try to make $x^4+y^4+z^4$ from these equations.
I did this:
$(x^2+y^2+z^2)^2=x^4+y^4+z^4+2x^2y^2+2x^2z^2+2y^2z^2$
And from here I can see:
$x^4+y^4+z^4=1-2(x^2y^2+x^2z^2+y^2z^2)$
But I'm missing $(x^2y^2+x^2z^2+y^2z^2)$
I tried finding $(x+y+z)^2$ and other combinations of squaring, tried multiplying some of these, but still couldn't find what I needed.
I'm in high school so try and do this for my level of knowledge, with at least a similar way to mine

Alternativley, working with trinomials is cumbersome, so make them binomials:
$$\begin{cases}x+y+z=1\\x^2+y^2+z^2={3\over2}\\x^3+y^3+z^3=1\end{cases} \Rightarrow \begin{cases}x+y=1-z\\x^2+y^2={3\over2}-z^2\\x^3+y^3=1-z^3\end{cases}.$$
$(1)^2-(2)$:
$$2xy=2z^2-2z-\frac12$$
$(1)^3-(1)$:
$$3xy(x+y)=3z(z-1) \Rightarrow 3xy(1-z)=3z(z-1) \Rightarrow xy=-z$$
So:
$$-2z=2z^2-2z-\frac12 \Rightarrow z=\pm \frac12.$$
$(1)^4$:
$$(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4 \Rightarrow \\
x^4+y^4=(x+y)^4-4xy(x^2+y^2)-6(xy)^2 \Rightarrow \\
x^4+y^4+z^4=(1-z)^4-4(-z)(\frac32-z^2)-6(-z)^2+z^4.$$
Plugging $z=\frac12$:
$$x^4+y^4+z^4=\frac1{16}+\frac52-\frac32+\frac1{16}=\frac98.$$
Plugging $z=-\frac12$:
$$x^4+y^4+z^4=\frac{81}{16}-\frac52-\frac32+\frac1{16}=\frac98.$$
