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.

x + y + z = 5
On squaring both sides,
$(x + y + z)^2 = 25$
$x^2 + y^2 + z^2 + 2xy + 2yz + 2zx = 25$
$21 + 2xy + 2yz + 2zx = 25$
$2xy + 2yz + 2zx = 25 - 21$
$2xy + 2yz + 2zx = 4$
$xy + yz + zx = 2$
Also,
$x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)$
Putting values,
$80 - 3xyz = (5)\left[21 - (xy + yz + zx)\right]$
80 - 3xyz = (5)(21 - 2)
80 - 3xyz = 95
-3xyz = 15
xyz = -5
