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HMMT February

Algebra Difficulty 5.2 AIME, harder Find the answer

Given that a,b,ca, b, c are positive real numbers and log⁡ab+log⁡bc+log⁡ca=0\log _{a} b+\log _{b} c+\log _{c} a=0, find the value of (log⁡ab)3+(log⁡bc)3+(log⁡ca)3\left(\log _{a} b\right)^{3}+\left(\log _{b} c\right)^{3}+\left(\log _{c} a\right)^{3}.

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

3. Let x=log⁡abx=\log _{a} b and y=log⁡bcy=\log _{b} c; then log⁡ca=−(x+y)\log _{c} a=-(x+y). Thus we want to compute the value of x3+y3−(x+y)3=−3x2y−3xy2=−3xy(x+y)x^{3}+y^{3}-(x+y)^{3}=-3 x^{2} y-3 x y^{2}=-3 x y(x+y). On the other hand, −xy(x+y)=(log⁡ab)(log⁡bc)(log⁡ca)=1-x y(x+y)=\left(\log _{a} b\right)\left(\log _{b} c\right)\left(\log _{c} a\right)=1, so the answer is 3.

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