Question 200367
{{{a^(1/3)+b^(1/3)+c^(1/3)=0}}}
{{{a^(1/3)+b^(1/3)=-c^(1/3)}}}
{{{(a^(1/3)+b^(1/3))^3=(-c^(1/3))^3}}} cube both sides
{{{a+3a^(2/3)b^(1/3)+3a^(1/3)b^(2/3)+b=-c}}}
{{{a+b+3a^(1/3)b^(1/3)(a^(1/3)+b^(1/3))=-c}}}
{{{a+b+c+3a^(1/3)b^(1/3)(a^(1/3)+b^(1/3))=0}}}
Noting that {{{a^(1/3)+b^(1/3)=-c^(1/3)}}}, we have
{{{a+b+c+3a^(1/3)b^(1/3)(-c^(1/3))=0}}}
{{{a+b+c=3a^(1/3)b^(1/3)c^(1/3)}}}
{{{(a+b+c)/3=a^(1/3)b^(1/3)c^(1/3)}}}
So
log[(a+b+c)/3]=1/3[loga+logb+logc]