4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ this is not so easy to solve! $\endgroup$ – Dr. Sonnhard Graubner Jan 14 '17 at 14:15
  • $\begingroup$ Look up Newton-Girard. $\endgroup$ – J. M. is a poor mathematician Jan 14 '17 at 15:19
  • $\begingroup$ Sage tells me that $x,y,z$ are the three roots of $x^3 - 5x^2 + 2x + 5 = 0$, but I'm not sure this would make for an interesting answer. $\endgroup$ – MvG Jan 14 '17 at 15:59
21
$\begingroup$

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 $xyz=-5$.

$\endgroup$
  • $\begingroup$ this is insane.. $\endgroup$ – Euler_Salter Feb 25 '17 at 15:27
7
$\begingroup$

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

$\endgroup$
4
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.