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Question Number 140040 by bobhans last updated on 03/May/21

((45+29(√2)))^(1/3) + ((45−29(√2)))^(1/3) =?

$$\:\sqrt[{\mathrm{3}}]{\mathrm{45}+\mathrm{29}\sqrt{\mathrm{2}}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{45}−\mathrm{29}\sqrt{\mathrm{2}}}\:=?\: \\ $$

Answered by cherokeesay last updated on 05/May/21

U = ((45 +29(√2)))^(1/3) V = ((45−29(√2)))^(1/3) U^( 3) = 45 +29(√2) V^( 3) = 45 −29(√2) U^( 3) + V^( 3) = 90 (U+V)^3 = U^( 3) + V^( 3) + 3UV (U + V) UV = ((45^2 −29^2 .2))^(1/3) = ((343))^(1/3) = 7 (U+V)^3 = 90 + 3×7(U + V) U+V = X X^( 3) −21X −90 = 0 ⇔ (X − 6)(X^2 + 6X +15) = 0 ⇒ X = 6 ((45 + 29(√2)))^(1/3) − ((45 − 29(√2)))^(1/3) = 6

$${U}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{45}\:+\mathrm{29}\sqrt{\mathrm{2}}}\:\:\:\:\:{V}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{45}−\mathrm{29}\sqrt{\mathrm{2}}} \\ $$$${U}^{\:\mathrm{3}} \:=\:\mathrm{45}\:+\mathrm{29}\sqrt{\mathrm{2}}\:\:\:\:{V}^{\:\mathrm{3}} \:=\:\mathrm{45}\:−\mathrm{29}\sqrt{\mathrm{2}}\:\: \\ $$$${U}^{\:\mathrm{3}} \:+\:{V}^{\:\:\mathrm{3}} \:=\:\mathrm{90} \\ $$$$\left({U}+{V}\right)^{\mathrm{3}} \:=\:{U}^{\:\mathrm{3}} \:+\:{V}^{\:\:\mathrm{3}} \:+\:\mathrm{3}{UV}\:\left({U}\:+\:{V}\right) \\ $$$${UV}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{45}^{\mathrm{2}} \:−\mathrm{29}^{\mathrm{2}} .\mathrm{2}}\:=\:\sqrt[{\mathrm{3}}]{\mathrm{343}}\:=\:\mathrm{7} \\ $$$$\left({U}+{V}\right)^{\mathrm{3}} \:=\:\mathrm{90}\:+\:\mathrm{3}×\mathrm{7}\left({U}\:+\:{V}\right)\:\:\:\:\:{U}+{V}\:=\:{X} \\ $$$${X}^{\:\mathrm{3}} \:−\mathrm{21}{X}\:−\mathrm{90}\:=\:\mathrm{0}\:\Leftrightarrow\: \\ $$$$\left({X}\:−\:\mathrm{6}\right)\left({X}^{\mathrm{2}} \:+\:\mathrm{6}{X}\:+\mathrm{15}\right)\:=\:\mathrm{0}\:\Rightarrow\:{X}\:=\:\mathrm{6} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{45}\:+\:\mathrm{29}\sqrt{\mathrm{2}}}\:−\:\sqrt[{\mathrm{3}}]{\mathrm{45}\:−\:\mathrm{29}\sqrt{\mathrm{2}}}\:=\:\mathrm{6} \\ $$$$ \\ $$

Answered by abdullahoudou last updated on 03/May/21

(3+(√2))+(3−(√2))=6

$$\left(\mathrm{3}+\sqrt{\mathrm{2}}\right)+\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)=\mathrm{6} \\ $$$$ \\ $$

Answered by MJS_new last updated on 03/May/21

generally: a^(1/3) +b^(1/3) =x a+3a^(1/3) b^(1/3) (a^(1/3) +b^(1/3) )+b=x^3 3a^(1/3) b^(1/3) x=x^3 −a−b 27abx^3 =(x^3 −a−b)^3 a=u+v∧b=u−v 27(u^2 −v^2 )x^3 =(x^3 −2u)^3 ... x^9 −6ux^6 −3(5u^2 −9v^2 )x^3 −8u^3 =0 u=45∧v=29(√2) x^9 −270x^6 +15039x^3 −729000=0 x=((t+90))^(1/3) t^3 −9261t−833490=0 (t−126)(t^2 +126t+6615)=0 t=126 x=6

$$\mathrm{generally}: \\ $$$${a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} ={x} \\ $$$${a}+\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} \left({a}^{\mathrm{1}/\mathrm{3}} +{b}^{\mathrm{1}/\mathrm{3}} \right)+{b}={x}^{\mathrm{3}} \\ $$$$\mathrm{3}{a}^{\mathrm{1}/\mathrm{3}} {b}^{\mathrm{1}/\mathrm{3}} {x}={x}^{\mathrm{3}} −{a}−{b} \\ $$$$\mathrm{27}{abx}^{\mathrm{3}} =\left({x}^{\mathrm{3}} −{a}−{b}\right)^{\mathrm{3}} \\ $$$${a}={u}+{v}\wedge{b}={u}−{v} \\ $$$$\mathrm{27}\left({u}^{\mathrm{2}} −{v}^{\mathrm{2}} \right){x}^{\mathrm{3}} =\left({x}^{\mathrm{3}} −\mathrm{2}{u}\right)^{\mathrm{3}} \\ $$$$... \\ $$$${x}^{\mathrm{9}} −\mathrm{6}{ux}^{\mathrm{6}} −\mathrm{3}\left(\mathrm{5}{u}^{\mathrm{2}} −\mathrm{9}{v}^{\mathrm{2}} \right){x}^{\mathrm{3}} −\mathrm{8}{u}^{\mathrm{3}} =\mathrm{0} \\ $$$${u}=\mathrm{45}\wedge{v}=\mathrm{29}\sqrt{\mathrm{2}} \\ $$$${x}^{\mathrm{9}} −\mathrm{270}{x}^{\mathrm{6}} +\mathrm{15039}{x}^{\mathrm{3}} −\mathrm{729000}=\mathrm{0} \\ $$$${x}=\sqrt[{\mathrm{3}}]{{t}+\mathrm{90}} \\ $$$${t}^{\mathrm{3}} −\mathrm{9261}{t}−\mathrm{833490}=\mathrm{0} \\ $$$$\left({t}−\mathrm{126}\right)\left({t}^{\mathrm{2}} +\mathrm{126}{t}+\mathrm{6615}\right)=\mathrm{0} \\ $$$${t}=\mathrm{126} \\ $$$${x}=\mathrm{6} \\ $$

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