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avatar+292 

Given that

\((x+y+z)(xy+xz+yz)=18\)

and that

\(x^2(y+z)+y^2(x+z)+z^2(x+y)=6\)

for real numbers x, y, and z, what is the value of xyz?

 

 

-hihihi

 

😎😎😎

 Jan 12, 2021
 #1
avatar+292 
+2

(x+y+z)(xy+xz+yz)=25,  x^2(y+z)+y^2(x+z)+z^2(x+y)=6

 

Expand the first equation 

 

yx^2 + zx^2 + xyz  + xy^2 + xyz + zy^2  + xyz + xz^2 + yz^2   =  18  simplify

 

yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2 + 3xyz = 6     (1) 

 

Expand the second equation

 

yx^2 + zx^2 + xy^2 + zy^2 + xz^2 + yz^2  = 6      (2)

 

Substitute  (2) into (1)

 

6 + 3xyz  =   18       subtract 6 from 18

 

3xyz   = 12            divide both sides by 3

 

xyz  = 4

 

-hihihi

 

😎😎😎

 Jan 12, 2021
 #2
avatar+428 
+2

You answered your own question surprise

 Jan 12, 2021
 #3
avatar+292 
+1

Sorry abbout that I found the anwer after I send the question so I just answered it

 Jan 12, 2021
edited by hihihi  Jan 12, 2021

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