Question Number 206322 by MATHEMATICSAM last updated on 12/Apr/24
$$\mathrm{If}\:{abc}\:=\:\mathrm{1}\:\mathrm{then}\:\mathrm{prove}\:\mathrm{that} \\ $$$$\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{a}\:+\:{b}^{−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{b}\:+\:{c}^{−\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\mathrm{1}\:+\:{c}\:+\:{a}^{−\mathrm{1}} }\:=\:\mathrm{1} \\ $$
Commented by Rasheed.Sindhi last updated on 12/Apr/24
$${For}\:{similar}\:{question}\:{see} \\ $$$${Q}#\mathrm{205380} \\ $$
Answered by som(math1967) last updated on 12/Apr/24
![(b/(b+ab+1))+(1/(1+b+ab))+((ab)/(ab+abc+b)) [ abc=1 ∴ab=c^(−1) ] =(b/(b+1+ab))+(1/(b+1+ab)) +((ab)/(b+1+ab)) =((b+1+ab)/(b+1+ab))=1](https://www.tinkutara.com/question/Q206326.png)
$$\frac{{b}}{{b}+{ab}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{1}+{b}+{ab}}+\frac{{ab}}{{ab}+{abc}+{b}} \\ $$$$\left[\:{abc}=\mathrm{1}\:\therefore{ab}={c}^{−\mathrm{1}} \right] \\ $$$$=\frac{{b}}{{b}+\mathrm{1}+{ab}}+\frac{\mathrm{1}}{{b}+\mathrm{1}+{ab}}\:+\frac{{ab}}{{b}+\mathrm{1}+{ab}} \\ $$$$=\frac{{b}+\mathrm{1}+{ab}}{{b}+\mathrm{1}+{ab}}=\mathrm{1} \\ $$