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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-

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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)
$$\mathrm{Find}\:\mathrm{a}\:\mathrm{triple}\:\mathrm{of}\:\mathrm{rational}\: \\ $$$$\mathrm{numbers}\:\left(\mathrm{a},\mathrm{b},\mathrm{c}\right)\:\mathrm{such}\:\mathrm{that}\: \\ $$$$\:\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{a}}\:+\sqrt[{\mathrm{3}}]{\mathrm{b}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{c}}\: \\ $$
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)..
$${t}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\Rightarrow\:{t}^{\mathrm{3}} =\mathrm{2}\:\rightarrow\:{t}^{\mathrm{3}} −\mathrm{1}=\mathrm{1} \\ $$$${t}−\mathrm{1}=\frac{{t}^{\mathrm{3}} −\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}\:=\:\frac{\mathrm{1}}{{t}^{\mathrm{2}} +{t}+\mathrm{1}}\:=\:\frac{\mathrm{3}}{\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{3}} \\ $$$$=\frac{\mathrm{3}}{{t}^{\mathrm{3}} +\mathrm{3}{t}^{\mathrm{2}} +\mathrm{3}{t}+\mathrm{1}}\:=\:\frac{\mathrm{3}}{\left({t}+\mathrm{1}\right)^{\mathrm{3}} } \\ $$$$\sqrt[{\mathrm{3}}]{{t}−\mathrm{1}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}}{{t}+\mathrm{1}}\:;\: \\ $$$${t}^{\mathrm{3}} +\mathrm{1}=\mathrm{3}\:\:\:\:\:,\:\:\:\:\:\: \\ $$$$\sqrt[{\mathrm{3}}]{{t}−\mathrm{1}}=\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{{t}^{\mathrm{3}} +\mathrm{1}}\:=\:\frac{\sqrt[{\mathrm{3}}]{\mathrm{3}}\left({t}^{\mathrm{2}} −{t}+\mathrm{1}\right)}{\mathrm{3}}=\frac{{t}^{\mathrm{2}} −{t}+\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{9}}} \\ $$$${t}=\sqrt[{\mathrm{3}}]{\mathrm{2}}\:\Rightarrow\:\sqrt[{\mathrm{3}}]{\sqrt[{\mathrm{3}}]{\mathrm{2}}−\mathrm{1}}=\:\sqrt[{\mathrm{3}}]{\frac{\mathrm{4}}{\mathrm{9}}}+\sqrt[{\mathrm{3}}]{−\frac{\mathrm{2}}{\mathrm{9}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{1}}{\mathrm{9}}} \\ $$$$\therefore\because\:\:{a}=\frac{\mathrm{4}}{\mathrm{9}}\:\:,\:\:{b}=−\frac{\mathrm{2}}{\mathrm{9}}\:\:,\:\:{c}=\frac{\mathrm{1}}{\mathrm{9}}.. \\ $$
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
$$\mathbb{N}\:=\:\left\{\mathrm{1},\mathrm{2},\mathrm{3},…\right\} \\ $$$$\boldsymbol{{i}}\:=\:\sqrt{−\mathrm{1}} \\ $$$$\mathbb{C}\:=\:{Set}\:{of}\:{Complex}\:{Numbers} \\ $$$$\boldsymbol{{e}}\:=\:{Base}\:{of}\:{Natural}\:{Logarithm} \\ $$$$!\:=\:{Symbol}\:{for}\:{factorial} \\ $$

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