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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.

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  • $\begingroup$ this is not so easy to solve! $\endgroup$ Jan 14, 2017 at 14:15
  • $\begingroup$ Look up Newton-Girard. $\endgroup$ Jan 14, 2017 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, 2017 at 15:59

4 Answers 4

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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 $xyz=-5$.

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  • $\begingroup$ this is insane.. $\endgroup$ Feb 25, 2017 at 15:27
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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

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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.

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\begin{align} x + z + y &= u=5 \tag{1}\label{1} ,\\ x^2 + z^2 + y^2 &= v=21 \tag{2}\label{2},\\ x^3 +z^3 + y^3 &= w=80 \tag{3}\label{3}. \end{align}

What is the value of $xyz$?

Surprisingly, the Ravi substitution works in this case, despite that not all the numbers $x,y,z$ are positive, and hence, the corresponding triangle is "unreal".

So, let \begin{align} a &= y + z ,\quad b = z + x ,\quad c = x + y \tag{4}\label{4} ,\\ x&=\rho-a ,\quad y=\rho-b ,\quad z=\rho-c \tag{5}\label{5} , \end{align}

where the triplet $a, b, c$ represents the sides of a triangle with semiperimeter $\rho$, inradius $r$ and circumradius $R$.

Then

\begin{align} x + z + y &= \rho \tag{6}\label{6} ,\\ x^2 + z^2 + y^2 &= \rho^2-2(r^2+4rR) \tag{7}\label{7} ,\\ x^3 + z^3 + y^3 &= \rho(\rho^2-12rR) \tag{8}\label{8} ,\\ xyz&=\rho\,r^2 \tag{9}\label{9} . \end{align}

Excluding $rR$ from \eqref{7}-\eqref{8}, we get

\begin{align} r^2&= \tfrac16\,\rho^2-\tfrac12\,v+\tfrac13\,\frac w{\rho} \tag{10}\label{10} , \end{align}

and from \eqref{9} we have the answer

\begin{align} xyz&= \tfrac16\,\rho^3-\tfrac12\,v\rho+\tfrac13\,w = \tfrac16\,5^3-\tfrac12\,21\cdot5+\tfrac13\,80 =-5 \tag{11}\label{11} . \end{align}

As a bonus, we can find that

\begin{align} rR &= \tfrac1{12}\,\frac{\rho^3-w}{\rho} \tag{12}\label{12} \end{align}

and $x=\rho-a,\ y=\rho-b,\ z=\rho-c$ are the roots of cubic equation

\begin{align} x^3-\rho\,x^2+(r^2+4rR)\,x-\rho r^2&=0 \tag{13}\label{13} ,\\ \text{or }\quad x^3-\rho\,x^2+\tfrac12\,(\rho^2-v)\,x-\tfrac13\,w-\tfrac16\,\rho\,(\rho^2-3v)&=0 \tag{14}\label{14} ,\\ x^3-5\,x^2+2\,x+5&=0 \tag{15}\label{15} . \end{align}

One of the roots of \eqref{15} is

\begin{align} x &= \tfrac53+ \tfrac23\,\sqrt{19}\,\cos\Big(\tfrac13\,\arctan(\tfrac9{25}\,\sqrt{331})\Big) \approx 4.253418 \tag{16}\label{16} ,\\ \text{the other two are }\quad y&\approx -0.773387 ,\quad z\approx 1.519969 \tag{17}\label{17} . \end{align}

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