Question Number 199015 by Tawa11 last updated on 26/Oct/23
Answered by Rasheed.Sindhi last updated on 26/Oct/23
$$\sqrt{{a}−{x}}\:+\sqrt{{b}−{x}}\:+\sqrt{{c}−{x}}\:=\mathrm{0} \\ $$$$\sqrt{{a}−{x}}\:=\mathrm{0}\:\wedge\:\sqrt{{b}−{x}}\:=\mathrm{0}\:\wedge\:\sqrt{{c}−{x}}\:=\mathrm{0} \\ $$$${x}={a}={b}={c} \\ $$$$\left({a}+{b}+{c}+\mathrm{3}{x}\right)\left({a}+{b}+{c}−{x}\right)=\mathrm{4}\left({bc}+{ca}+{ab}\right) \\ $$$$\left({x}+{x}+{x}+\mathrm{3}{x}\right)\left({x}+{x}+{x}−{x}\right)=\mathrm{4}\left({x}.{x}+{x}.{x}+{x}.{x}\right) \\ $$$$\left(\mathrm{6}{x}\right)\left(\mathrm{2}{x}\right)=\mathrm{4}\left(\mathrm{3}{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:\mathrm{12}{x}^{\mathrm{2}} =\mathrm{12}{x}^{\mathrm{2}} \:\left({proved}\right) \\ $$
Commented by Tawa11 last updated on 26/Oct/23
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/23
$$\mathrm{Show}\:\mathrm{that}:\:\:\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:+\sqrt[{\mathrm{3}}]{{c}}\:=\mathrm{0}\Rightarrow\left({a}+{b}+{c}\right)^{\mathrm{3}} =\mathrm{27}{abc} \\ $$$${Let}\:\sqrt[{\mathrm{3}}]{{a}}\:={A}\:,\:\sqrt[{\mathrm{3}}]{{b}}\:={B}\:,\:\sqrt[{\mathrm{3}}]{{c}}\:={C}\Rightarrow{A}+{B}+{C}=\mathrm{0} \\ $$$$\:{A}^{\mathrm{3}} +{B}^{\mathrm{3}} +{C}^{\mathrm{3}} −\mathrm{3}{ABC}=\left({A}+{B}+{C}\right)\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} −{AB}−{BC}−{CA}\right) \\ $$$$\:{A}^{\mathrm{3}} +{B}^{\mathrm{3}} +{C}^{\mathrm{3}} −\mathrm{3}{ABC}=\mathrm{0} \\ $$$$\:{A}^{\mathrm{3}} +{B}^{\mathrm{3}} +{C}^{\mathrm{3}} \:=\:\mathrm{3}{ABC} \\ $$$$\:\:{a}+{b}+{c}=\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\: \\ $$$$\:\left({a}+{b}+{c}\right)^{\mathrm{3}} =\left(\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\right)^{\mathrm{3}} =\mathrm{27}{abc} \\ $$$$\:\:\: \\ $$
Commented by Tawa11 last updated on 26/Oct/23
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$