回復 6# satsuki931000 的帖子
選擇第一題:
設\(\displaystyle a=\root 3\of {\frac{3-\sqrt{5}}{2}}+\root 3 \of{\frac{3+\sqrt{5}}{2}}\),則\(a^6-6a^4+9a^2+27\)之值為
(A)34 (B)36 (C)38 (D)40
[解答]
利用 \(\left(x+y\right)^3 = x^3+y^3+3xy\left(x+y\right)\),可得
\(\displaystyle \left(\sqrt[3]{\frac{3+\sqrt{5}}{2}}+\sqrt[3]{\frac{3-\sqrt{5}}{2}}\right)^3 = \frac{3+\sqrt{5}}{2}+\frac{3-\sqrt{5}}{2}+3\cdot\sqrt[3]{\frac{3+\sqrt{5}}{2}}\cdot \sqrt[3]{\frac{3-\sqrt{5}}{2}}\left(\sqrt[3]{\frac{3+\sqrt{5}}{2}}+\sqrt[3]{\frac{3-\sqrt{5}}{2}}\right)\)
\(\Rightarrow a^3 =3+3a\)
\(\Rightarrow a^3-3a=3\)
所以,\(\displaystyle a^6-6a^4+9a^2+27 = \left(a^3-3a\right)^2+27 = 36\)