Question Number 207919 by efronzo1 last updated on 30/May/24
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Commented by Frix last updated on 30/May/24
$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0}\vee\mathrm{15} \\ $$
Answered by mr W last updated on 30/May/24
$${a}={b}={c}=\mathrm{0}\:{is}\:{a}\:{solution} \\ $$$${for}\:{a}\neq\mathrm{0},\:{b}\neq\mathrm{0},\:{c}\neq\mathrm{0}: \\ $$$$\left({ii}\right)×\left({iii}\right): \\ $$$${bc}={bc}\left({a}+\mathrm{2}\right)\left({a}−\mathrm{2}\right) \\ $$$$\Rightarrow\left({a}+\mathrm{2}\right)\left({a}−\mathrm{2}\right)=\mathrm{1}\:\Rightarrow{a}^{\mathrm{2}} =\mathrm{5} \\ $$$${a}={b}^{\mathrm{2}} \left({a}−\mathrm{2}\right) \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\frac{{a}}{{a}−\mathrm{2}}={a}\left({a}+\mathrm{2}\right)={a}^{\mathrm{2}} +\mathrm{2}{a} \\ $$$${c}^{\mathrm{2}} ={b}^{\mathrm{2}} \left({a}−\mathrm{2}\right)^{\mathrm{2}} ={a}\left({a}−\mathrm{2}\right)={a}^{\mathrm{2}} −\mathrm{2}{a} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{a}+{a}^{\mathrm{2}} −\mathrm{2}{a}=\mathrm{3}{a}^{\mathrm{2}} \\ $$$$\:\:\:\:=\mathrm{15} \\ $$$$\Rightarrow{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{0}\:{or}\:\mathrm{15} \\ $$