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Question

If x2+y2+z2=2(xyz)3, find the value of 2x3y+4z.

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Solution

Given, x2+y2+z2=2(xyz)3
(x22x)+(y2+2y)+(z2+2z)+3=0
(x22x+1)+(y2+2y+1)+(z2+2z+1)=0
(x1)2+(y+1)2+(z+1)2=0
(x1)=0 and (y+1)=0 and (z+1)=0
x=1 and y=1 and z=1
2x3y+4z=2+34=1

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