If x, y, z beolg to R, logxy, logzx, logyz are in ap
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According to question,
logxylogzx=logzxlogyz⇒(logx)3=(logz)3
⇒x=z
Since 2y3=x3+z3⇒x2=y3 or x=y
Given xyz=64 & x=y=z
∴x=y=z=4
& x+y+z=12
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