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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)

a−x+b−x+c−x=0a−x=0∧b−x=0∧c−x=0x=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(3x2)12x2=12x2(proved)

Commented by Tawa11 last updated on 26/Oct/23

Thanks sir. I appreciate.

Thankssir.Iappreciate.

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

Showthat:a3+b3+c3=0⇒(a+b+c)3=27abcLeta3=A,b3=B,c3=C⇒A+B+C=0A3+B3+C3−3ABC=(A+B+C)(A2+B2+C2−AB−BC−CA)A3+B3+C3−3ABC=0A3+B3+C3=3ABCa+b+c=3abc3(a+b+c)3=(3abc3)3=27abc

Commented by Tawa11 last updated on 26/Oct/23

God bless you sir. Thanks for your time.

Godblessyousir.Thanksforyourtime.

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