Question Number 196185 by cortano12 last updated on 19/Aug/23
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Commented by AST last updated on 19/Aug/23
$${When}\:{x}={y}={z}\:\wedge\:{a}={b}={c};\frac{{ab}+{bc}+{ca}}{{bc}}=\mathrm{3} \\ $$$${When}\:{x}=−\mathrm{1},{y}=\mathrm{1},{z}=\mathrm{1};\:\frac{{ab}+{bc}+{ca}}{{bc}}=\frac{{a}\left({b}+{c}\right)}{{bc}}+\mathrm{1} \\ $$$${can}\:{get}\:{multiple}\:{values}\:{since}\:{a},{b},{c}\:{can}\:{vary}. \\ $$
Answered by AST last updated on 19/Aug/23
$${xyz}={x}^{\mathrm{3}} \Rightarrow{k}^{\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{{ab}+{bc}+{ca}}{{abc}}\right)} ={k}^{\frac{\mathrm{1}}{\mathrm{8}{a}^{\mathrm{3}} }} \Rightarrow\frac{{ab}+{bc}+{ca}}{{bc}}=\frac{\mathrm{1}}{\mathrm{4}{a}^{\mathrm{2}} } \\ $$$${When}\:{x}={y}={z}\:\wedge\:{a}={b}={c};\frac{{ab}+{bc}+{ca}}{{bc}}=\mathrm{3} \\ $$$${When}\:{x}=−\mathrm{1},{y}=\mathrm{1},{z}=\mathrm{1};\:\frac{{ab}+{bc}+{ca}}{{bc}}=\frac{{a}\left({b}+{c}\right)}{{bc}}+\mathrm{1} \\ $$$${can}\:{get}\:{multiple}\:{values}\:{since}\:{a},{b},{c}\:{can}\:{vary}. \\ $$
Answered by horsebrand11 last updated on 19/Aug/23
$$\:\Rightarrow\mathrm{x}^{\mathrm{2}} \:=\:\mathrm{y}^{\frac{\mathrm{2b}}{\mathrm{a}}} \:\mathrm{and}\:\mathrm{z}=\mathrm{y}^{\frac{\mathrm{b}}{\mathrm{c}}} \\ $$$$\:\:\mathrm{then}\:\mathrm{y}^{\frac{\mathrm{2b}}{\mathrm{a}}} \:=\:\mathrm{y}^{\mathrm{1}+\frac{\mathrm{b}}{\mathrm{c}}} \\ $$$$\:\:\frac{\mathrm{2b}}{\mathrm{a}}\:=\:\frac{\mathrm{b}+\mathrm{c}}{\mathrm{c}}\Rightarrow\:\mathrm{ab}+\mathrm{ca}\:=\:\mathrm{2bc} \\ $$$$\:\mathrm{so}\:\frac{\mathrm{ab}+\mathrm{bc}+\mathrm{ca}}{\mathrm{bc}}\:=\:\frac{\mathrm{3bc}}{\mathrm{bc}}\:=\:\mathrm{3} \\ $$