Question Number 217385 by Rasheed.Sindhi last updated on 12/Mar/25
$${a},{b},{c}\in\mathbb{R} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{9} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{21} \\ $$$${ab}+{bc}+{ca}=? \\ $$
Commented by mr W last updated on 12/Mar/25
$${i}\:{think}\:{even}\:{for}\:{real}\:{numbers}\:{there} \\ $$$${is}\:{no}\:{unique}\:{solution}. \\ $$
Commented by mr W last updated on 12/Mar/25
$${say}\:{x}={a}+{b}+{c},\:{y}={ab}+{bc}+{ca},\:{z}={abc} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{9}+\mathrm{2}{y}\:\:\:…\left({i}\right) \\ $$$$\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)\left({a}+{b}+{c}\right)={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\left({a}+{b}+{c}\right)\left({ab}+{bc}+{ca}\right)−\mathrm{3}{abc} \\ $$$$\Rightarrow\mathrm{9}{x}=\mathrm{21}+{xy}−\mathrm{3}{z}\:\:\:…\left({ii}\right) \\ $$$$\left({i}\right)\:{and}\:\left({ii}\right)\:{has}\:{infinite}\:{solutions}! \\ $$
Commented by Rasheed.Sindhi last updated on 12/Mar/25
$$\mathcal{T}{hanks}\:\boldsymbol{{sir}}\:\boldsymbol{{mr}}\:\boldsymbol{{W}}. \\ $$