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.

We have $$(x+y+z)^3=(x^3+y^3+z^3)+3x(y^2+z^2)+3y(x^2+z^2)+3z(x^2+y^2)+6xyz.$$ Hence
$$125=80+3x(21-x^2)+3y(21-y^2)+3z(21-z^2)+6xyz.$$
This leads to
$$45=63(x+y+z)-3(x^3+y^3+z^3)+6xyz.$$
This gives us $45=315-240+6xyz$, so $6xyz=-30$ and $xyz=-5$.
