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Question Number 149883 by liberty last updated on 08/Aug/21
Prove that ((2+(√5)))^(1/3) +((2−(√5)))^(1/3)  is  a rational number
$$\mathrm{Prove}\:\mathrm{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:\mathrm{is} \\ $$$$\mathrm{a}\:\mathrm{rational}\:\mathrm{number} \\ $$
Answered by john_santu last updated on 08/Aug/21
Let x=((2+(√5)))^(1/3)  +((2−(√5)))^(1/3)    We then have x−((2+(√5)))^(1/3) −((2−(√5)))^(1/3)  =0  as we have seen a+b+c = 0 implies  a^3 +b^3 +c^3 =3abc so we obtain  x^3 −(2+(√5))−(2−(√5))=3x(((2+(√5))(2−(√5))))^(1/3)   or x^3 +3x−4=0 . clearly one of  the roots of this equation is x=1  and the other two roots satisfy  the equation x^2 +x+4=0 which  has no real solutions. since  ((2+(√5)))^(1/3)  +((2−(√5)))^(1/3)  is a real root  it follows that ((2+(√5)))^(1/3)  +((2−(√5)))^(1/3)  =1  which a rational number
$$\mathrm{L}{et}\:{x}=\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\: \\ $$$${We}\:{then}\:{have}\:{x}−\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}−\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{0} \\ $$$${as}\:{we}\:{have}\:{seen}\:{a}+{b}+{c}\:=\:\mathrm{0}\:{implies} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc}\:{so}\:{we}\:{obtain} \\ $$$${x}^{\mathrm{3}} −\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)−\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)=\mathrm{3}{x}\sqrt[{\mathrm{3}}]{\left(\mathrm{2}+\sqrt{\mathrm{5}}\right)\left(\mathrm{2}−\sqrt{\mathrm{5}}\right)} \\ $$$${or}\:{x}^{\mathrm{3}} +\mathrm{3}{x}−\mathrm{4}=\mathrm{0}\:.\:{clearly}\:{one}\:{of} \\ $$$${the}\:{roots}\:{of}\:{this}\:{equation}\:{is}\:{x}=\mathrm{1} \\ $$$${and}\:{the}\:{other}\:{two}\:{roots}\:{satisfy} \\ $$$${the}\:{equation}\:{x}^{\mathrm{2}} +{x}+\mathrm{4}=\mathrm{0}\:{which} \\ $$$${has}\:{no}\:{real}\:{solutions}.\:{since} \\ $$$$\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:{is}\:{a}\:{real}\:{root} \\ $$$${it}\:{follows}\:{that}\:\sqrt[{\mathrm{3}}]{\mathrm{2}+\sqrt{\mathrm{5}}}\:+\sqrt[{\mathrm{3}}]{\mathrm{2}−\sqrt{\mathrm{5}}}\:=\mathrm{1} \\ $$$${which}\:{a}\:{rational}\:{number} \\ $$

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