Let a,b,c be positive real numbers such that logab+logbc+logca=0.Find (logab)3+(logbc)3+(logca)3.
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Let x=logab,y=logbc, and z=logca. Then x+y+z=0, so x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−xz−yz)=0.Hence, x3+y3+z3=3xyz=3(logab)(logbc)(logca)=3⋅logalogb⋅logblogc⋅logcloga=3.