题号:3371    题型:解答题    来源:2000年全国初中数学竞赛试题 入库日期 2022/12/9 11:05:59
已知$a=\sqrt{3+\sqrt{8}}$, 求$a^3+a^{-3}$
【答案】 解:
$$
\begin{aligned}
a & =\sqrt{3+2 \sqrt{2}} \\
& =\sqrt{\sqrt{2}+1^2+2 \sqrt{2} \times 1} \\
& =\sqrt{(\sqrt{2}+1)^2} \\
& =\sqrt{2}+1
\end{aligned}
$$
所以
$$
a^{-1}=\dfrac{1}{a}= \Rightarrow \frac{1}{a}=\sqrt{2}-1
$$
$$
\begin{aligned}
& a^3+\frac{1}{a^3} \\
= & \left(a+\frac{1}{a}\right)\left(a^2-1+\frac{1}{a^2}\right) \\
= & \left(a+\frac{1}{a}\right)\left[\left(a+\frac{1}{a}\right)^2-2-1\right] \\
= & 2 \sqrt{2} \cdot\left[(2 \sqrt{2})^2-3\right] \\
= & 2 \sqrt{2} \times 5=10 \sqrt{2} .
\end{aligned}
$$


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