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Question Number 141153 by iloveisrael last updated on 16/May/21

Answered by iloveisrael last updated on 16/May/21

Commented by SamuelAsaana last updated on 16/May/21

i donot underdtand sir please explain for for me

idonotunderdtandsirpleaseexplainforforme

Answered by mr W last updated on 16/May/21

(x−y)^3 =5((√3)−(√2)) ⇒x−y=((5((√3)−(√2))))^(1/3) (x+y)^3 −6xy(x+y)=5((√3)+(√2)) (x+y)^3 −(3/2)((x+y)^2 −(x−y)^2 )(x+y)=5((√3)+(√2)) (x+y)^3 −3((25(5−2(√6))))^(1/3) (x+y)+10((√3)+(√2))=0 Δ=−25(5−2(√6))+25(5+2(√6))=100(√6)>0 ⇒x+y=((10(6)^(1/4) −5((√3)+(√2))))^(1/3) −((10(6)^(1/4) +5((√3)+(√2)))^(1/3) ) x=(1/2)[((10(6)^(1/4) −5((√3)+(√2))))^(1/3) −((10(6)^(1/4) +5((√3)+(√2))))^(1/3) +((5((√3)−(√2))))^(1/3) ]≈−1.209505 y=(1/2)[((10(6)^(1/4) −5((√3)+(√2))))^(1/3) −((10(6)^(1/4) +5((√3)+(√2))))^(1/3) −((5((√3)−(√2))))^(1/3) ]≈−2.376472

(x−y)3=5(3−2)⇒x−y=5(3−2)3(x+y)3−6xy(x+y)=5(3+2)(x+y)3−32((x+y)2−(x−y)2)(x+y)=5(3+2)(x+y)3−325(5−26)3(x+y)+10(3+2)=0Δ=−25(5−26)+25(5+26)=1006>0⇒x+y=1064−5(3+2)3−1064+5(3+23)x=12[1064−5(3+2)3−1064+5(3+2)3+5(3−2)3]≈−1.209505y=12[1064−5(3+2)3−1064+5(3+2)3−5(3−2)3]≈−2.376472

Commented by iloveisrael last updated on 16/May/21

Yes

Yes

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