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Question Number 190093 by mnjuly1970 last updated on 27/Mar/23

$$$$$$provethat:$$$$$$$$c={(\sqrt{5}+2)}^{\frac{1}{3}}-{(\sqrt{5}-2)}^{\frac{1}{3}}$$$$$$$$\mathrm{is}\mathrm{a}rational\mathrm{number}.$$$$$$$$$$

Commented by MJS_new last updated on 27/Mar/23

$$\mathrm{I}{\mathrm{don}}^{\prime }\mathrm{t}\mathrm{think}\mathrm{it}\mathrm{is}\mathrm{rational}$$

Commented by som(math1967) last updated on 18/May/23

$$yessir,ithinkitisnotrational$$

Commented by mnjuly1970 last updated on 27/Mar/23

$$yessirthanksalot$$$$icorretedit.\sqrt{5}insteadof\sqrt{10}$$$$$$

Answered by MJS_new last updated on 27/Mar/23

$$...={(2+\sqrt{5})}^{1/3}-{(2+\sqrt{5})}^{-1/3}$$$${(2+\sqrt{5})}^{1/3}=\frac{1+\sqrt{5}}{2}$$$$\Rightarrow c=1$$

Answered by som(math1967) last updated on 27/Mar/23

$$c^3=(\sqrt{5}+2)-(\sqrt{5}-2)-3{(5-4)}^{\frac{1}{3}}.c$$$$[c={(\sqrt{5}+2)}^{\frac{1}{3}}-{(\sqrt{5}-2)}^{\frac{1}{3}}]$$$$c^3=4-3c$$$$c^3+3c-4=0$$$$c^3-1+3(c-1)=0$$$$(c-1)(c^2+c+1+3)=0$$$$(c-1)(c^2+c+4)=0$$$$forrealvalueofc(c^2+c+4)\ne 0$$$$c=1ratinalnumber$$

Commented by mehdee42 last updated on 27/Mar/23

$$thatwasperfect.$$

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