System of 3 equations problem

Question

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

Answer 1

Hint:

$$(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$

$$(x+y+z)^2=?$$

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

Answer 2

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 $z=-\frac12$: $$x^4+y^4+z^4=\frac{81}{16}-\frac52-\frac32+\frac1{16}=\frac98.$$