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Question-199015




Question Number 199015 by Tawa11 last updated on 26/Oct/23
Answered by Rasheed.Sindhi last updated on 26/Oct/23
(√(a−x)) +(√(b−x)) +(√(c−x)) =0  (√(a−x)) =0 ∧ (√(b−x)) =0 ∧ (√(c−x)) =0  x=a=b=c  (a+b+c+3x)(a+b+c−x)=4(bc+ca+ab)  (x+x+x+3x)(x+x+x−x)=4(x.x+x.x+x.x)  (6x)(2x)=4(3x^2 )     12x^2 =12x^2  (proved)
$$\sqrt{{a}−{x}}\:+\sqrt{{b}−{x}}\:+\sqrt{{c}−{x}}\:=\mathrm{0} \\ $$$$\sqrt{{a}−{x}}\:=\mathrm{0}\:\wedge\:\sqrt{{b}−{x}}\:=\mathrm{0}\:\wedge\:\sqrt{{c}−{x}}\:=\mathrm{0} \\ $$$${x}={a}={b}={c} \\ $$$$\left({a}+{b}+{c}+\mathrm{3}{x}\right)\left({a}+{b}+{c}−{x}\right)=\mathrm{4}\left({bc}+{ca}+{ab}\right) \\ $$$$\left({x}+{x}+{x}+\mathrm{3}{x}\right)\left({x}+{x}+{x}−{x}\right)=\mathrm{4}\left({x}.{x}+{x}.{x}+{x}.{x}\right) \\ $$$$\left(\mathrm{6}{x}\right)\left(\mathrm{2}{x}\right)=\mathrm{4}\left(\mathrm{3}{x}^{\mathrm{2}} \right) \\ $$$$\:\:\:\mathrm{12}{x}^{\mathrm{2}} =\mathrm{12}{x}^{\mathrm{2}} \:\left({proved}\right) \\ $$
Commented by Tawa11 last updated on 26/Oct/23
Thanks sir. I appreciate.
$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$
Answered by Rasheed.Sindhi last updated on 26/Oct/23
Show that:  (a)^(1/3)  +(b)^(1/3)  +(c)^(1/3)  =0⇒(a+b+c)^3 =27abc  Let (a)^(1/3)  =A , (b)^(1/3)  =B , (c)^(1/3)  =C⇒A+B+C=0   A^3 +B^3 +C^3 −3ABC=(A+B+C)(A^2 +B^2 +C^2 −AB−BC−CA)   A^3 +B^3 +C^3 −3ABC=0   A^3 +B^3 +C^3  = 3ABC    a+b+c=3((abc))^(1/3)     (a+b+c)^3 =(3((abc))^(1/3) )^3 =27abc
$$\mathrm{Show}\:\mathrm{that}:\:\:\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:+\sqrt[{\mathrm{3}}]{{c}}\:=\mathrm{0}\Rightarrow\left({a}+{b}+{c}\right)^{\mathrm{3}} =\mathrm{27}{abc} \\ $$$${Let}\:\sqrt[{\mathrm{3}}]{{a}}\:={A}\:,\:\sqrt[{\mathrm{3}}]{{b}}\:={B}\:,\:\sqrt[{\mathrm{3}}]{{c}}\:={C}\Rightarrow{A}+{B}+{C}=\mathrm{0} \\ $$$$\:{A}^{\mathrm{3}} +{B}^{\mathrm{3}} +{C}^{\mathrm{3}} −\mathrm{3}{ABC}=\left({A}+{B}+{C}\right)\left({A}^{\mathrm{2}} +{B}^{\mathrm{2}} +{C}^{\mathrm{2}} −{AB}−{BC}−{CA}\right) \\ $$$$\:{A}^{\mathrm{3}} +{B}^{\mathrm{3}} +{C}^{\mathrm{3}} −\mathrm{3}{ABC}=\mathrm{0} \\ $$$$\:{A}^{\mathrm{3}} +{B}^{\mathrm{3}} +{C}^{\mathrm{3}} \:=\:\mathrm{3}{ABC} \\ $$$$\:\:{a}+{b}+{c}=\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\: \\ $$$$\:\left({a}+{b}+{c}\right)^{\mathrm{3}} =\left(\mathrm{3}\sqrt[{\mathrm{3}}]{{abc}}\right)^{\mathrm{3}} =\mathrm{27}{abc} \\ $$$$\:\:\: \\ $$
Commented by Tawa11 last updated on 26/Oct/23
God bless you sir.  Thanks for your time.
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\ $$$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}. \\ $$

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