SOLUTION: If a^(1/3)+b^(1/3)+c^(1/3)=0, then show that log[(a+b+c)/3]=1/3[loga+logb+logc]

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Question 200367: If a^(1/3)+b^(1/3)+c^(1/3)=0, then show that log[(a+b+c)/3]=1/3[loga+logb+logc]
Answer by orca(409) About Me  (Show Source):
You can put this solution on YOUR website!
a%5E%281%2F3%29%2Bb%5E%281%2F3%29%2Bc%5E%281%2F3%29=0
a%5E%281%2F3%29%2Bb%5E%281%2F3%29=-c%5E%281%2F3%29
%28a%5E%281%2F3%29%2Bb%5E%281%2F3%29%29%5E3=%28-c%5E%281%2F3%29%29%5E3 cube both sides
a%2B3a%5E%282%2F3%29b%5E%281%2F3%29%2B3a%5E%281%2F3%29b%5E%282%2F3%29%2Bb=-c
a%2Bb%2B3a%5E%281%2F3%29b%5E%281%2F3%29%28a%5E%281%2F3%29%2Bb%5E%281%2F3%29%29=-c
a%2Bb%2Bc%2B3a%5E%281%2F3%29b%5E%281%2F3%29%28a%5E%281%2F3%29%2Bb%5E%281%2F3%29%29=0
Noting that a%5E%281%2F3%29%2Bb%5E%281%2F3%29=-c%5E%281%2F3%29, we have
a%2Bb%2Bc%2B3a%5E%281%2F3%29b%5E%281%2F3%29%28-c%5E%281%2F3%29%29=0
a%2Bb%2Bc=3a%5E%281%2F3%29b%5E%281%2F3%29c%5E%281%2F3%29
%28a%2Bb%2Bc%29%2F3=a%5E%281%2F3%29b%5E%281%2F3%29c%5E%281%2F3%29
So
log[(a+b+c)/3]=1/3[loga+logb+logc]