Is this Algebra?

$\large \begin{aligned} (x+y+z)(x^2+y^2+z^2-xy-yz-xz) &=x^3+y^3+z^3-3xyz \\ \Rightarrow \left(\sqrt{x^2+y^2+z^2+2(xy+yz+xz)}\right)(x^2+y^2+z^2) &=1 \\ \Rightarrow \left(\sqrt{x^2+y^2+z^2}\right)(x^2+y^2+z^2) &=1 \\ \Rightarrow (x^2+y^2+z^2)(x^2+y^2+z^2)^2 &=1 \\ \Rightarrow (x^2+y^2+z^2)^3 &=1 \\ \Rightarrow x^2+y^2+z^2 &= \boxed{1} \end{aligned}$

I know this is simple algebra manipulation, but I will present a different solution.

Observe that $\begin{vmatrix} x & y & z\\ z & x & y\\ y & z & x \end{vmatrix} = x^3 + y^3 + z^3 - 3xyz = 1$

However, this is just the volume of the parallelepiped bounded by the vectors $\begin{pmatrix} x \\ y \\ z \end{pmatrix}, \begin{pmatrix} z \\ x \\ y \end{pmatrix}, \begin{pmatrix} y \\ z \\ x \end{pmatrix}$ . Notice that they have equal magnitude. However, since $xy + yz + zx = 0$ , we know that the pairwise dot product of the three vectors is 0, which implies that this parallelepiped is a cube. Thus, since the side length is $\sqrt{x^2 + y^2 + z^2}$ , we can write the volume as $(x^2 + y^2 + z^2)^{\frac{3}{2}} = 1 \implies x^2 + y^2 + z^2 = \boxed{1}$

let: $s_1=x+y+z,s_2=xy+yz+zx=0,s_3=xyz,p_n=x^n+y^n+z^n$ . by newtons sum $\begin{aligned} p_1=s_1\\ p_2=s_1p_1-2s_2=s_1^2\\ p_3=s_1p_2-s_2p_1+3s_3\rightarrow p_3-3s_3=s_1(s_1^2)-0p_1=s_1^3\end{aligned}$ note the the expression given in question is $p_3-3s_3=1$ : $s_1^3=1\rightarrow s_1 = 1(unreal-neglected)$ so, $p_2=x^2+y^2+z^2=s_1^2=\boxed{1}$