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Question Number 216694 by sniper237 last updated on 16/Feb/25
Prove that ^3 (√((√5)+2)) −^3 (√((√5)−2)) =1
Provethat35+2−35−2=1
Answered by golsendro last updated on 16/Feb/25
let x= (((√5)+2))^(1/3) −(((√5)−2))^(1/3) (((√5)+2))^(1/3) −(((√5)−2))^(1/3) −x = 0 (√5) +2 −(√5) +2 −x^3 = 3x ((5−4))^(1/3) x^3 +3x−4 = 0 x=1
letx=5+23−5−235+23−5−23−x=05+2−5+2−x3=3x5−43x3+3x−4=0x=1
Answered by Ghisom last updated on 16/Feb/25
Φ=(1/2)+((√5)/2) Φ^(−1) =−(1/2)+((√5)/2) Φ^3 =2+(√5) Φ^(−3) =−2+(√5) the rest is easy
Φ=12+52Φ−1=−12+52Φ3=2+5Φ−3=−2+5therestiseasy
Answered by Rasheed.Sindhi last updated on 16/Feb/25
(((√5) +2))^(1/3) −(((√5) −2))^(1/3) =1 let (((√5) +2))^(1/3) =a (((√5) −2))^(1/3) =1/a Asume a−(1/a)=x a^3 −(1/a^3 )−3(a−(1/a))=x^3 ((√5) +2)−((√5) −2)−3(x)=x^3 x^3 +3x−4=0 (x−1)(x^2 +x+4)=0 ⇒x=1
5+23−5−23=1let5+23=a5−23=1/aAsumea−1a=xa3−1a3−3(a−1a)=x3(5+2)−(5−2)−3(x)=x3x3+3x−4=0(x−1)(x2+x+4)=0⇒x=1
Answered by Rasheed.Sindhi last updated on 16/Feb/25
(((√5) +2))^(1/3) −(((√5) −2))^(1/3) =1 let (((√5) +2))^(1/3) −(((√5) −2))^(1/3) =x (((√5) +2))^(1/3) −(((√5) −2))^(1/3) −x=0 −((−(√5) −2))^(1/3) −(((√5) −2))^(1/3) −x=0 ((−(√5) −2))^(1/3) +(((√5) −2))^(1/3) +x=0 determinant (((a+b+c=0_(⇒a^3 +b^3 +c^3 =3abc) ))) ⇒(−(√5) −2)+((√5) −2)+x^3 =3( ((−(√5) −2))^(1/3) )((((√5) −2))^(1/3) )(x) x^3 −4=−3x((((√5) +2)((√5)−2)))^(1/3) x^3 +3x(1)−4=0 (x−1)(x^2 +x+4)=0 ⇒x=1(proved)
5+23−5−23=1let5+23−5−23=x5+23−5−23−x=0−−5−23−5−23−x=0−5−23+5−23+x=0a+b+c=0⇒a3+b3+c3=3abc⇒(−5−2)+(5−2)+x3=3(−5−23)(5−23)(x)x3−4=−3x(5+2)(5−2)3x3+3x(1)−4=0(x−1)(x2+x+4)=0⇒x=1(proved)

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