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.

Consider the polynomial
$$p(t) = (1-x t)(1-y t)(1-z t)$$
Let's consider the series expansion of $\log\left[p(t)\right]$:
$$\log\left[p(t)\right] =-\sum_{k=1}^{\infty}\frac{S_k}{k} t^k$$
where
$$S_k = x^k + y^k + z^k$$
Since we're given the $S_k$ for $k$ up to $3$ we can write down the series expansion of $\log\left[p(t)\right]$ up to third order in $t$, but that's sufficient to calculate $p(t)$, as it's a third degree polynomial. The coefficient of $t^3$ equals $-xyz$, so we only need to focus on that term. We have:
$$\log\left[p(t)\right] = -\left(5 t +\frac{21}{2} t^2 + \frac{80}{3} t^3+\cdots\right)$$
Exponentiating yields:
$$p(t) = \exp(-5t)\exp\left(-\frac{21}{2}t^2\right)\exp\left(-\frac{80}{3}t^3\right)\times\exp\left[\mathcal{O}(t^4)\right]$$
The $t^3$ term can come in its entirety from the first factor, or we can pick the linear term in $t$ from there and then multiply that by the $t^2$ term from the second factor or we can take the $t^3$ term from the last factor. Adding up the 3 possibilities yields:
$$xyz = \frac{5^3}{3!} -5\times\frac{21}{2} + \frac{80}{3} = -5$$
It is also easy to show that $x^4 + y^4 + z^4 = 333$ by using that the coefficient of the $t^4$ term is zero.
