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 $xyz=-5$.
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.