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Question Number 119802 by bemath last updated on 27/Oct/20

$$Givena,b,crealnumberandnotequalto1.$$$$If\log {}_a\left(b\right)+\log {}_b\left(c\right)+\log {}_c\left(a\right)=0then$$$${\left(\log {}_a\left(b\right)\right)}^3+{\left(\log {}_b\left(c\right)\right)}^3+{\left(\log {}_c\left(a\right)\right)}^3=?$$

Answered by $@y@m last updated on 27/Oct/20

$$Ifl+m+n=0$$$$then{l}^3+m^3+n^3=3lmn$$$$Herel=\log {}_a\left(b\right)$$$$m=\log {}_b\left(c\right)$$$$n=\log {}_c\left(a\right)$$$$\therefore 3lmn=3\times \log {}_a\left(b\right)\times \log {}_b\left(c\right)\times \log {}_c\left(a\right)=3$$

Answered by Ar Brandon last updated on 27/Oct/20

$${\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{b}}\mathrm{c}+{\log }_{\mathrm{c}}\mathrm{a}=0$$$$\Rightarrow {({\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{b}}\mathrm{c})}^3=-{\left({\log }_{\mathrm{c}}\mathrm{a}\right)}^3$$$$\Rightarrow {\left({\log }_{\mathrm{a}}\mathrm{b}\right)}^3+3{\left({\log }_{\mathrm{a}}\mathrm{b}\right)}^2\left({\log }_{\mathrm{b}}\mathrm{c}\right)$$$$+3\left({\log }_{\mathrm{a}}\mathrm{b}\right){\left({\log }_{\mathrm{b}}\mathrm{c}\right)}^2+{\left({\log }_{\mathrm{b}}\mathrm{c}\right)}^3=-{\left({\log }_{\mathrm{c}}\mathrm{a}\right)}^3$$$$\Rightarrow {\left({\log }_{\mathrm{c}}\mathrm{a}\right)}^3+{\left({\log }_{\mathrm{a}}\mathrm{b}\right)}^3+{\left({\log }_{\mathrm{b}}\mathrm{c}\right)}^3$$$$=-3\left({\log }_{\mathrm{a}}\mathrm{b}\right)\left({\log }_{\mathrm{b}}\mathrm{c}\right)[{\log }_{\mathrm{a}}\mathrm{b}+{\log }_{\mathrm{b}}\mathrm{c}]$$$$=3\left({\log }_{\mathrm{c}}\mathrm{a}\right)\left({\log }_{\mathrm{a}}\mathrm{b}\right)\left({\log }_{\mathrm{b}}\mathrm{c}\right)$$$$=3\left(\frac{1}{{\log }_{\mathrm{a}}\mathrm{c}}\right)\left(\frac{{\log }_{\mathrm{a}}\mathrm{b}}{1}\right)\left(\frac{{\log }_{\mathrm{a}}\mathrm{c}}{{\log }_{\mathrm{a}}\mathrm{b}}\right)=3$$

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