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Question Number 170478 by balirampatel last updated on 24/May/22

(( 9 + 4(√5) ))^(1/3) + (( 9 − 4(√5) ))^(1/3) = ?

9+453+9−453=?

Answered by Rasheed.Sindhi last updated on 24/May/22

(( 9 + 4(√5) ))^(1/3) + (( 9 − 4(√5) ))^(1/3) =x(say);x∈R (( 9 + 4(√5) ))^(1/3) =a (1/a)=(1/( (( 9 + 4(√5) ))^(1/3) ))∙(((( 9 − 4(√5) ))^(1/3) )/( (( 9 − 4(√5) ))^(1/3) )) =(((( 9 − 4(√5) ))^(1/3) )/( ((81−16(5)))^(1/3) ))=(( 9 − 4(√5) ))^(1/3) =(( 9 + 4(√5) ))^(1/3) +(1/( (( 9 + 4(√5) ))^(1/3) ))=x =a+(1/a)=x a^3 +(1/a^3 )=9+4(√5) +9−4(√5)=18 (a+(1/a))^3 −3(a+(1/a))=18 (a+(1/a))^3 −3(a+(1/a))−18=0 x^3 −3x−18=0 (x−3)(x^2 +3x+6)=0 x−3=0 ∣ x^2 +3x+6=0(x∉R) Rejected x=3 (( 9 + 4(√5) ))^(1/3) + (( 9 − 4(√5) ))^(1/3) =3

9+453+9−453=x(say);x∈R9+453=a1a=19+453⋅9−4539−453=9−45381−16(5)3=9−453=9+453+19+453=x=a+1a=xa3+1a3=9+45+9−45=18(a+1a)3−3(a+1a)=18(a+1a)3−3(a+1a)−18=0x3−3x−18=0(x−3)(x2+3x+6)=0x−3=0∣x2+3x+6=0(x∉R)Rejectedx=39+453+9−453=3

Answered by mr W last updated on 24/May/22

(( 4(√5)+9 ))^(1/3) − (( 4(√5)−9 ))^(1/3) =(( (√((−((18)/2))^2 +(−(3/3))^3 ))+((18)/2) ))^(1/3) − (( (√((−((18)/2))^2 +(−(3/3))^3 ))−((18)/2) ))^(1/3) which is the real root of x^3 −3x−18=0 (x−3)(x^2 +3x+6)=0 ⇒x=3

45+93−45−93=(−182)2+(−33)3+1823−(−182)2+(−33)3−1823whichistherealrootofx3−3x−18=0(x−3)(x2+3x+6)=0⇒x=3

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