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x^2 + y^2 + z^2 = 18

xy + xz + yz = -7

 

Let x, y, and z be integers satisfying the system of equations above. Find xyz.

 Jul 6, 2021
 #1
avatar+795 
+1

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

$(x+y+z)^2 - 2(-7) = 18$

$(x+y+z)^2 = 4$

$x+y+z = 2$

 

$x^2 y^2 + 2 x^2 y z + x^2 z^2 + 2 x y^2 z + 2 x y z^2 + y^2 z^2 = 49$

$x^2 y^2 + x^2 z^2 + y^2 z^2 + 2xyz(x + y + z) = 49$

$\frac{xyz^2}{z^2} + \frac{xyz^2}{y^2} + \frac{xyz^2}{x^2} + 4xyz = 49$

$xyz \left(\frac{xyz}{z^2} + \frac{xyz}{y^2} + \frac{xyz}{x^2} \right) + 4xyz = 49$

$xyz \left(\frac{xyz(x^2 y^2 + y^2 z^2 +x^2 z^2)}{xyz^2} \right) + 4xyz = 49$

$xyz \left(\frac{18}{xyz} \right) + 4xyz = 49$

$18 + 4xyz = 49$

$4xyz = 31$

$xyz = \boxed{\frac{31}{4}}$

 #2
avatar+795 
+1

I made a mistake with the $\frac{18}{xyz}$, though I am scared to edit the latex : Do you think you can manipulate $x^2 y^2 + y^2 z^2 + x^2 z^2$ into something you already know? The answer follows from there.


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