Question Number 44729 by behi83417@gmail.com last updated on 03/Oct/18
Answered by MJS last updated on 04/Oct/18
$$\mathrm{trivial}\:\mathrm{solutions}\:{a}={b}={c}=\mathrm{0}\:\vee\:{a}={b}={c}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${c}^{\mathrm{2}} +{bc}+{b}^{\mathrm{2}} −{a}=\mathrm{0} \\ $$$${c}^{\mathrm{2}} +{ac}+{a}^{\mathrm{2}} −{b}=\mathrm{0} \\ $$$$\left({b}−{a}\right){c}+{b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{b}−{a}=\mathrm{0} \\ $$$${c}=−\left({a}+{b}+\mathrm{1}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab}+{a}+{b}+\mathrm{1}=\mathrm{0} \\ $$$${b}^{\mathrm{2}} +\left({a}+\mathrm{1}\right){b}+{a}^{\mathrm{2}} +{a}+\mathrm{1}=\mathrm{0} \\ $$$${b}=−\frac{{a}+\mathrm{1}}{\mathrm{2}}\pm\frac{\sqrt{−\mathrm{3}{a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{3}}}{\mathrm{2}} \\ $$$${c}=−\frac{{a}+\mathrm{1}}{\mathrm{2}}\mp\frac{\sqrt{−\mathrm{3}{a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{3}}}{\mathrm{2}} \\ $$$${a}={b}^{\mathrm{2}} +{c}^{\mathrm{2}} +{bc}\:\mathrm{always}\:\mathrm{true} \\ $$$${a}\in\mathbb{R}\:\Rightarrow\:{b},\:{c}\notin\mathbb{R} \\ $$
Answered by ajfour last updated on 04/Oct/18
$$\mathrm{3}{x}^{\mathrm{2}} ={x} \\ $$$$\Rightarrow\:\:\:{a}={b}={c}\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\:\:\:\left({x}\:{above}\right). \\ $$
Answered by ajfour last updated on 04/Oct/18
$${b}={c} \\ $$$${a}=\mathrm{3}{b}^{\mathrm{2}} \\ $$$${b}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{ab} \\ $$$${b}=\mathrm{9}{b}^{\mathrm{4}} +{b}^{\mathrm{2}} +\mathrm{3}{b}^{\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{9}{b}^{\mathrm{3}} +\mathrm{3}{b}^{\mathrm{2}} +{b}\:=\:\mathrm{1} \\ $$$$\Rightarrow\:{b}={c}=\:\frac{\mathrm{1}}{\mathrm{3}}\:=\:{a}\:. \\ $$
Commented by behi83417@gmail.com last updated on 04/Oct/18
$${thank}\:{you}\:{very}\:{much}\:{dear}\:{my} \\ $$$${friends}. \\ $$$$….{also}\:{waiting}\:{for}\:{mrW}\mathrm{3}'{s}\:{proof}. \\ $$
Answered by MrW3 last updated on 04/Oct/18
$$\left({i}\right)−\left({ii}\right): \\ $$$${a}−{b}={b}^{\mathrm{2}} −{a}^{\mathrm{2}} +{bc}−{ac} \\ $$$$\left({b}+{a}\right)\left({b}−{a}\right)+\left({b}−{a}\right){c}+\left({b}−{a}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}+{c}+\mathrm{1}\right)\left({b}−{a}\right)=\mathrm{0} \\ $$$${similarly} \\ $$$$\Rightarrow\left({a}+{b}+{c}+\mathrm{1}\right)\left({c}−{b}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({a}+{b}+{c}+\mathrm{1}\right)\left({a}−{c}\right)=\mathrm{0} \\ $$$$ \\ $$$${case}\:\mathrm{1}:\:{a}={b}={c}={k} \\ $$$$\Rightarrow{k}=\mathrm{3}{k}^{\mathrm{2}} \\ $$$$\Rightarrow{k}=\mathrm{0}\:{or}\:{k}=\frac{\mathrm{1}}{\mathrm{3}} \\ $$$$ \\ $$$${case}\:\mathrm{2}:\:{a}+{b}+{c}+\mathrm{1}=\mathrm{0} \\ $$$${a}=\left({b}+{c}\right)^{\mathrm{2}} −{bc}\:\:\:\:…\left(\mathrm{1}\right) \\ $$$${b}=\left({c}+{a}\right)^{\mathrm{2}} −{ca}\:\:\:\:…\left(\mathrm{2}\right) \\ $$$${c}=\left({a}+{b}\right)^{\mathrm{2}} −{ab}\:\:\:\:…\left(\mathrm{3}\right) \\ $$$$ \\ $$$${a}=\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −{bc} \\ $$$${b}=\left({a}+{c}\right)^{\mathrm{2}} −{ca} \\ $$$${c}=\left(\mathrm{1}+{c}\right)^{\mathrm{2}} −{ab} \\ $$$$ \\ $$$${a}=\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −{c}\left[\left({a}+{c}\right)^{\mathrm{2}} −{ca}\right] \\ $$$${a}=\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −{c}\left({a}+{c}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} {a} \\ $$$${a}^{\mathrm{2}} ={a}\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −{ac}\left({a}+{c}\right)^{\mathrm{2}} +{c}^{\mathrm{2}} {a}^{\mathrm{2}} \\ $$$${c}=\left(\mathrm{1}+{c}\right)^{\mathrm{2}} −{a}\left[\left({a}+{c}\right)^{\mathrm{2}} −{ca}\right] \\ $$$${c}=\left(\mathrm{1}+{c}\right)^{\mathrm{2}} −{a}\left({a}+{c}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} {c} \\ $$$${c}^{\mathrm{2}} ={c}\left(\mathrm{1}+{c}\right)^{\mathrm{2}} −{ac}\left({a}+{c}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −{c}^{\mathrm{2}} ={a}\left(\mathrm{1}+{a}\right)^{\mathrm{2}} −{c}\left(\mathrm{1}+{c}\right)^{\mathrm{2}} \\ $$$${a}^{\mathrm{2}} −{c}^{\mathrm{2}} ={a}+\mathrm{2}{a}^{\mathrm{2}} +{a}^{\mathrm{3}} −{c}−\mathrm{2}{c}^{\mathrm{2}} −{c}^{\mathrm{3}} \\ $$$${a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} ={c}+{c}^{\mathrm{2}} +{c}^{\mathrm{3}} ={h}\:{say} \\ $$$${similarly} \\ $$$${a}+{a}^{\mathrm{2}} +{a}^{\mathrm{3}} ={b}+{b}^{\mathrm{2}} +{b}^{\mathrm{3}} ={h} \\ $$$${since}\:{x}+{x}^{\mathrm{2}} +{x}^{\mathrm{3}} ={h}\:{has}\:{only}\:{one}\:{single} \\ $$$${real}\:{solution},\:{lets}\:{say}\:{x}={p}, \\ $$$${we}\:{get}\:{a}={b}={c}={x}={p} \\ $$$${i}.{e}.\:{case}\:\mathrm{2}\:{gives}\:{the}\:{same}\:{solution}\:{as} \\ $$$${case}\:\mathrm{1}. \\ $$
Commented by behi83417@gmail.com last updated on 04/Oct/18
$${perfect}!\:{thank}\:{you}\:{dear}\:{master}. \\ $$