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+1
76
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avatar+33 

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 \(xyz\)?

 Dec 11, 2019
 #1
avatar
0

xyz = 14.

 Dec 11, 2019
 #2
avatar+106539 
+1

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

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

 

(x + y + z)(xy + xz + yz)   =  18

x^2y + x^2z + xyz + xy^2 + xyz + y^2z + xyz + xz^2 + yz^2   = 18

(x^2y  + x^2z + xy^2  + zy^2 + xz^2 + yz^2)   +  3xyz  =  18

(6)   +  3xyz    =  18

3xyz  =    12

xyz  =  4

 

 

cool cool cool

 Dec 11, 2019

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