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Just wanted to ask regarding finding the solutions for a $3$ variable simultaneous equation problem. It seems to be quite trivial, but I'm not sure how to approach simplifying a problem like this:

  • $xyz = 576$
  • $x + y + z = 26$
  • $xy + xz + yz = 216$

Thanks in advance for your help! I feel a little stupid that I can't wrap my head around this!

Sam Hoffman
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Mathsn00b
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    Do you know Vieta's formulas? If so, is there some polynomial that $x$, $y$, and $z$ are all roots of? – Joseph Camacho May 14 '22 at 15:12
  • Thanks for the quick response! Unfortunately, I'm not familiar with Vieta's formulas. This isn't derived from any polynomial in the question. This is essentially all the information that's provided. – Mathsn00b May 14 '22 at 15:15
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    Well then I'd advise looking up Vieta's formulas.

    1. https://en.wikipedia.org/wiki/Vieta%27s_formulas

    2. https://artofproblemsolving.com/wiki/index.php/Vieta%27s_formulas

    3. https://brilliant.org/wiki/vietas-formula/

    4. https://www.andrew.cmu.edu/user/daltizio/Vietas%20Formulas.pdf

     – Joseph Camacho May 14 '22 at 15:29

2 Answers2

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Here is one solution that doesn't make any use of Vieta's formulas but does require some basic algebra and solving quadratics:

$($$x$ $+$ $y$ $+$ $z$$)$$^2$ $=$ $($$x$ $+$ $($$y$ $+$ $z$$)$$)$$^2$

$=$ $x$$^2$ $+$ $($$y$ $+$ $z$$)$$^2$ $+$ $2x(y + z)$

$=$ $x$$^2$ $+$ $y$$^2$ $+$ $z$$^2$ $+$ $2yz$ $+$ $2x$$(y + z)$

$= x^2 + y^2 + z^2 + 2(xy + xz + yz)$

$(26)^2 = x^2 + y^2 + z^2 + 2(216)$

$(26)^2 - 2(216) = 576 - 432 = 144$

$y^2 + z^2 = 144 - x^2$

$yz = 576/x$

i.e. $(y + z)^2 = y^2 + z^2 + 2yz$

$(y + z)^2 = 144 - x^2 + 1152/x$

$(y + z) = 26 - x$

Hence $(26 - x)^2 = 144 - x^2 + 1152/x$

$576 + x^2 - 52x = 144 - x^2 + 1152/x$

$2x^2 - 52x + 432 = 1152/x$

$2x^2 - 52x + 432 - 1152/x = 0$

By the quadratic formula $x = (52 +-$ $\sqrt(52^2 - 8$.$(432 -1152/x))$$)$$/2$

$x = 26 +- $($\sqrt(-752 + 9216/x)$)$/2$

implying $9216/x > 752$ and $-752 + 9216/x$ is a square

$9216/752 > x$

$12.2553 > x$

$12 \ge x$

try $x = 12$ yes!

if $x = 12$, then $yz = 576/12 = 48$

$y + z = 26 - x = 14$

substitute $z = 48/y$ into $y + z = 14$

$y + 48/y = 14$

$y^2 - 14y + 48 = 0$

$y = (14 +-$ $\sqrt(14^2 - 4$.$48)$$)/2$

$y = 7 +- 1$

say $y = 6$ and $z = 8$ or vice versa

Joebloggs
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Substituting $z=26-x-y$ and taking the resultant of the remaining two polynomials in $x,y$ given by the system of polynomial equation, one obtains directly $$ (x,y,z)=(12,8,6) $$ and all permutations of it (the system is symmetric in $x,y,z$).

Without using resultants, the two remaining polynomial equations are $$ - x^2y - xy^2 + 26xy - 576 = 0, $$ and $$ - x^2 - xy + 26x - y^2 + 26y - 216 =0 $$ Multiplying the second equation by $y$ and substracting the first equation yields $$ (y - 6)(y - 8)(y - 12)=0. $$

Dietrich Burde
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