Răspuns:
[tex]\sqrt{3+\sqrt{5}}=\sqrt[6]{(3+\sqrt{5})^3}=\sqrt[6]{72+32\sqrt{5}}[/tex]
[tex]\sqrt[3]{\sqrt{5}-1}=\sqrt[6]{(\sqrt{5}-1)^2}=\sqrt[6]{6-2\sqrt{5}}[/tex]
Atunci
[tex]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[/tex]
Explicație pas cu pas:
