Find $xyz$ given that $x + z + y = 5$, $x^2 + z^2 + y^2 = 21$, $x^3 + z^3 + y^3 = 80$

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 $xyz=-5$.

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

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.