Question Number 21471 by Joel577 last updated on 24/Sep/17
$$\mathrm{If}\:\:{a}\:+\:{b}\:+\:{c}\:=\:\mathrm{0},\:\mathrm{then} \\ $$$$\frac{\left({a}\:+\:{b}\right)\left({b}\:+\:{c}\right)\left({a}\:+\:{c}\right)}{{abc}}\:\mathrm{is}\:\mathrm{equal}\:\mathrm{to}\:... \\ $$
Answered by Tinkutara last updated on 24/Sep/17
$${a}+{b}=−{c} \\ $$$$\therefore\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{{abc}}=\frac{\left(−{c}\right)\left(−{a}\right)\left(−{b}\right)}{{abc}}=−\mathrm{1} \\ $$
Commented by Joel577 last updated on 24/Sep/17
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by Rasheed.Sindhi last updated on 25/Sep/17
$$\underset{−} {\:\:\:\mathrm{An}\:\mathrm{other}\:\mathrm{way}\:\:\:}\:\: \\ $$$${a}+{b}+{c}=\mathrm{0}\Rightarrow\begin{cases}{{a}=−\left({b}+{c}\right)}\\{{b}=−\left({a}+{c}\right)}\\{{c}=−\left({a}+{b}\right)}\end{cases} \\ $$$$\frac{\left({a}\:+\:{b}\right)\left({b}\:+\:{c}\right)\left({a}\:+\:{c}\right)}{{abc}}= \\ $$$$\frac{\left({a}\:+\:{b}\right)\left({b}\:+\:{c}\right)\left({a}\:+\:{c}\right)}{−\left({b}+{c}\right).−\left({a}+{c}\right).−\left({a}+{b}\right)}=−\mathrm{1} \\ $$