SOLUTION: x,y,z are real number. if xyz=78 and x^2+y^2+z^2=206, determine x+y+z

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Question 165924: x,y,z are real number. if xyz=78 and x^2+y^2+z^2=206, determine x+y+z
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
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x,y,z are real number. if xyz = 78 and x^2+y^2+z^2=206, determine x+y+z
:
Assume that one of the variables = 1 (z)
:
Three factors of 78: x=6, y=13, z=1
Then
6^2 + 13^2 + 1^2 = 206
:
x+y+z sum:
6 + 13 + 1 = 20