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Question Number 152281 by john_santu last updated on 27/Aug/21

Find a triple of rational numbers (a,b,c) such that (((2)^(1/3) −1))^(1/3) = (a)^(1/3) +(b)^(1/3) + (c)^(1/3)

Findatripleofrationalnumbers(a,b,c)suchthat23−13=a3+b3+c3

Answered by puissant last updated on 27/Aug/21

t=(2)^(1/3) ⇒ t^3 =2 → t^3 −1=1 t−1=((t^3 −1)/(t^2 +t+1)) = (1/(t^2 +t+1)) = (3/(3t^2 +3t+3)) =(3/(t^3 +3t^2 +3t+1)) = (3/((t+1)^3 )) ((t−1))^(1/3) =((3)^(1/3) /(t+1)) ; t^3 +1=3 , ((t−1))^(1/3) =(((3)^(1/3) (t^2 −t+1))/(t^3 +1)) = (((3)^(1/3) (t^2 −t+1))/3)=((t^2 −t+1)/( (9)^(1/3) )) t=(2)^(1/3) ⇒ (((2)^(1/3) −1))^(1/3) = ((4/9))^(1/3) +((−(2/9)))^(1/3) +((1/9))^(1/3) ∴∵ a=(4/9) , b=−(2/9) , c=(1/9)..

t=23⇒t3=2→t3−1=1t−1=t3−1t2+t+1=1t2+t+1=33t2+3t+3=3t3+3t2+3t+1=3(t+1)3t−13=33t+1;t3+1=3,t−13=33(t2−t+1)t3+1=33(t2−t+1)3=t2−t+193t=23⇒23−13=493+−293+193∴∵a=49,b=−29,c=19..

Commented by Rasheed.Sindhi last updated on 27/Aug/21

N = {1,2,3,...} i = (√(−1)) C = Set of Complex Numbers e = Base of Natural Logarithm ! = Symbol for factorial

N={1,2,3,...}i=−1C=SetofComplexNumberse=BaseofNaturalLogarithm!=Symbolforfactorial

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