Question

Aratati ca numarul a=$\sqrt{3+\sqrt{5}$*$\sqrt[3]{\sqrt{5}-1 }$*$\sqrt[6]{7-3\sqrt{5} }$∈Q.

Answer 1

Răspuns:

$\sqrt{3+\sqrt{5}}=\sqrt[6]{(3+\sqrt{5})^3}=\sqrt[6]{72+32\sqrt{5}}$

$\sqrt[3]{\sqrt{5}-1}=\sqrt[6]{(\sqrt{5}-1)^2}=\sqrt[6]{6-2\sqrt{5}}$

Atunci

$a=\sqrt[6]{72+32\sqrt{5}}\cdot\sqrt[6]{6-2\sqrt{5}}\cdot\sqrt[6]{7-3\sqrt{5}}=\\=\sqrt[6]{(72+32\sqrt{5})(6-2\sqrt{5})(7-3\sqrt{5})}=\sqrt[6]{72+32\sqrt{5})(72-32\sqrt{5})}=\\=\sqrt[6]{72^2-5\cdot 32^2}=\sqrt[6]{5184-5120}=\sqrt[6]{64}=2$

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