Question Number 72027 by mathmax by abdo last updated on 23/Oct/19
$${if}\:{a}+{b}+{c}\:=\mathrm{3}\:\:{and}\:{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} \:+{c}^{\mathrm{3}} =\mathrm{27}\:{find}\:{a}^{\mathrm{2}} \:+{b}^{\mathrm{2}} \:+{c}^{\mathrm{2}} \\ $$
Commented by Rasheed.Sindhi last updated on 26/Oct/19
$${Insufficient}\:{data}\:{for}\:{unique}\:{solution}. \\ $$$${There}\:{are}\:{three}\:{unknowns},{so}\:{three} \\ $$$${equations}\:{are}\:{required}\:{for}\:{unique} \\ $$$${solution}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/19
$${For}\:{unique}\:{solution}\:{a}\:{third}\:{equation} \\ $$$${is}\:{necessary},{because}\:{there}\:{are}\:{three} \\ $$$${unknowns}. \\ $$$${Since}\:{the}\:{third}\:{equation}\:{is}\:{not}\:{given} \\ $$$${we}\:{assume}\:{value}\:{of}\:{one}\:{unknown}: \\ $$$${Let}\:{c}=\mathrm{1} \\ $$$${a}+{b}=\mathrm{2}\:,\:{a}^{\mathrm{3}} \:+{b}^{\mathrm{3}} \:=\mathrm{26} \\ $$$$\left({a}+{b}\right)^{\mathrm{3}} =\mathrm{8} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +\mathrm{3}{ab}\left({a}+{b}\right)=\mathrm{8} \\ $$$$\mathrm{26}+\mathrm{3}{ab}\left(\mathrm{2}\right)=\mathrm{8} \\ $$$${ab}=\frac{−\mathrm{18}}{\mathrm{6}}=−\mathrm{3} \\ $$$${a}\left(\mathrm{2}−{a}\right)=−\mathrm{3} \\ $$$${a}^{\mathrm{2}} −\mathrm{2}{a}−\mathrm{3}=\mathrm{0} \\ $$$$\left({a}−\mathrm{3}\right)\left({a}+\mathrm{1}\right)=\mathrm{0} \\ $$$${a}=\mathrm{3}\:\vee\:{a}=−\mathrm{1} \\ $$$${b}=\mathrm{2}−\mathrm{3}=−\mathrm{1}\:\vee\:{b}=\mathrm{2}+\mathrm{1}=\mathrm{3} \\ $$$$\left({a},{b},{c}\right)=\left(\mathrm{3},−\mathrm{1},\mathrm{1}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:{or} \\ $$$$\left({a},{b},{c}\right)=\left(−\mathrm{1},\mathrm{3},\mathrm{1}\right) \\ $$$$ \\ $$$${If}\:{c}=\mathrm{1},−\mathrm{1}\:{or}\:\mathrm{3}: \\ $$$$\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left(−\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{3}\right)^{\mathrm{2}} +\left(\mathrm{1}\right)^{\mathrm{2}} =\mathrm{6} \\ $$$$ \\ $$