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

Hint:
$$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$
$$(x+y+z)^2=?$$
$$x^3+y^3+z^3-3xyz=(x+y+z)((x^2+y^2+z^2-(xy+yz+zx))$$
