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

$$\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}=?$$

Answered by cherokeesay last updated on 05/May/21

$$U=\sqrt[3]{45+29\sqrt{2}}V=\sqrt[3]{45-29\sqrt{2}}$$$$U^3=45+29\sqrt{2}{V}^3=45-29\sqrt{2}$$$$U^3+V^3=90$$$${(U+V)}^3=U^3+V^3+3UV(U+V)$$$$UV=\sqrt[3]{{45}^2-{29}^2.2}=\sqrt[3]{343}=7$$$${(U+V)}^3=90+3\times 7(U+V)U+V=X$$$$X^3-21X-90=0\iff$$$$(X-6)(X^2+6X+15)=0\Rightarrow X=6$$$$\sqrt[3]{45+29\sqrt{2}}-\sqrt[3]{45-29\sqrt{2}}=6$$$$$$

Answered by abdullahoudou last updated on 03/May/21

$$(3+\sqrt{2})+(3-\sqrt{2})=6$$$$$$

Answered by MJS_new last updated on 03/May/21

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

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