Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$ I was looking back in my junk, then I found this:

$$x + z + y = 5$$
$$x^2 + z^2 + y^2 = 21$$
$$x^3 + z^3 + y^3 = 80$$
What is the value of $xyz$?
A) $5$
B) $4$
C) $1$
D) $-4$
E) $-5$

It's pretty easy, any chances of solving this question? I already have the
answer for this, but I didn't fully understand.
Thanks for the attention.


\begin{align} x + z + y &= u=5 \tag{1}\label{1} ,\\ x^2 + z^2 + y^2 &= v=21 \tag{2}\label{2},\\ x^3 +z^3 + y^3 &= w=80 \tag{3}\label{3}. \end{align}
What is the value of $xyz$?

Surprisingly, the Ravi substitution
works in this case, despite that
not all the numbers $x,y,z$ are positive,
and hence, the corresponding triangle is "unreal".
So, let
\begin{align} 
a &= y + z
,\quad
b = z + x
,\quad
c = x + y
\tag{4}\label{4}
,\\
x&=\rho-a
,\quad
y=\rho-b
,\quad
z=\rho-c
\tag{5}\label{5}
,
\end{align}
where the triplet $a, b, c$ represents the sides of a triangle
with semiperimeter $\rho$, inradius $r$ and circumradius $R$.
Then
\begin{align} 
x + z + y &= \rho
\tag{6}\label{6}
,\\
x^2 + z^2 + y^2 &= \rho^2-2(r^2+4rR)
\tag{7}\label{7}
,\\
x^3 + z^3 + y^3 &= \rho(\rho^2-12rR)
\tag{8}\label{8}
,\\
xyz&=\rho\,r^2
\tag{9}\label{9}
.
\end{align}
Excluding $rR$ from \eqref{7}-\eqref{8},
we get
\begin{align} 
r^2&=
\tfrac16\,\rho^2-\tfrac12\,v+\tfrac13\,\frac w{\rho}
\tag{10}\label{10}
,
\end{align}
and from \eqref{9} we have the answer
\begin{align} 
xyz&=
\tfrac16\,\rho^3-\tfrac12\,v\rho+\tfrac13\,w
=
\tfrac16\,5^3-\tfrac12\,21\cdot5+\tfrac13\,80
=-5
\tag{11}\label{11}
.
\end{align}
As a bonus, we can find that
\begin{align} 
rR &= \tfrac1{12}\,\frac{\rho^3-w}{\rho}
\tag{12}\label{12}
\end{align}
and $x=\rho-a,\ y=\rho-b,\ z=\rho-c$ are the roots of cubic equation
\begin{align}
x^3-\rho\,x^2+(r^2+4rR)\,x-\rho r^2&=0
\tag{13}\label{13}
,\\
\text{or }\quad
x^3-\rho\,x^2+\tfrac12\,(\rho^2-v)\,x-\tfrac13\,w-\tfrac16\,\rho\,(\rho^2-3v)&=0
\tag{14}\label{14}
,\\
x^3-5\,x^2+2\,x+5&=0
\tag{15}\label{15}
.
\end{align}
One of the roots of \eqref{15} is
\begin{align}
x &= \tfrac53+
\tfrac23\,\sqrt{19}\,\cos\Big(\tfrac13\,\arctan(\tfrac9{25}\,\sqrt{331})\Big)
\approx  4.253418
\tag{16}\label{16}
,\\
\text{the other two are }\quad
y&\approx -0.773387
,\quad
z\approx  1.519969
\tag{17}\label{17}
.
\end{align}
