Question Number 222003 by fantastic last updated on 14/Jun/25
$${If}\:{a}+{b}+{c}=\mathrm{0}\:{then}\:{prove}\:{that} \\ $$$$\frac{\mathrm{1}}{{x}^{{b}} +{x}^{−{c}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{{c}} +{x}^{−{a}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}}=\mathrm{1} \\ $$
Answered by som(math1967) last updated on 15/Jun/25
$$\:\frac{{x}^{{c}} }{{x}^{{b}+{c}} +{x}^{\mathrm{0}} +{x}^{{c}} }\:+\frac{\mathrm{1}}{{x}^{{c}} +{x}^{{b}+{c}} +\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:+\frac{{x}^{{b}+{c}} }{{x}^{{a}+{b}+{c}} +{x}^{{c}} +{x}^{{b}+{c}} }\:\:\bigstar \\ $$$$\frac{{x}^{{c}} }{{x}^{{c}} +\mathrm{1}+{x}^{{b}+{c}} }\:+\frac{\mathrm{1}}{{x}^{{c}} +\mathrm{1}+{x}^{{b}+{c}} }+\frac{{x}^{{b}+{c}} }{{x}^{{c}} +\mathrm{1}+{x}^{{b}+{c}} } \\ $$$$=\frac{{x}^{{c}} +\mathrm{1}+{x}^{{b}+{c}} }{{x}^{{c}} +\mathrm{1}+{x}^{{b}+{c}} }=\mathrm{1} \\ $$$$\bigstar\:{a}+{b}+{c}=\mathrm{0}\Rightarrow−{a}={b}+{c} \\ $$
Commented by fantastic last updated on 15/Jun/25
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Commented by Rasheed.Sindhi last updated on 15/Jun/25
$$\vee.\:\cap\boldsymbol{\mathrm{i}}\subset\in! \\ $$
Commented by som(math1967) last updated on 15/Jun/25
$${thank}\:{you} \\ $$
Answered by Rasheed.Sindhi last updated on 16/Jun/25
$$\mathrm{Another}\:\mathrm{way} \\ $$$${c}=−{a}−{b} \\ $$$$\frac{\mathrm{1}}{{x}^{{b}} +{x}^{{a}+{b}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{−{a}−{b}} +{x}^{−{a}} +\mathrm{1}}+\frac{\mathrm{1}}{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}} \\ $$$$=\frac{\mathrm{1}}{{x}^{{b}} +{x}^{{a}+{b}} +\mathrm{1}}+\frac{{x}^{{a}} }{{x}^{−{b}} +\mathrm{1}+{x}^{{a}} }+\frac{\mathrm{1}}{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}} \\ $$$$=\frac{{x}^{−{b}} }{\mathrm{1}+{x}^{{a}} +{x}^{−{b}} }+\frac{{x}^{{a}} }{{x}^{−{b}} +\mathrm{1}+{x}^{{a}} }+\frac{\mathrm{1}}{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}} \\ $$$$=\frac{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}}{{x}^{{a}} +{x}^{−{b}} +\mathrm{1}}=\mathrm{1}\left({Proved}\right) \\ $$$$ \\ $$