题号:3436    题型:填空题    来源:数学竞赛 入库日期 2022/12/14 8:17:06
若 $a=\sqrt[3]{36}+\sqrt[3]{30}+\sqrt[3]{25}$
则 $ a-\frac{1}{a^2}=$
【答案】 解: 令 $m=\sqrt[3]{6}, n=\sqrt[3]{5}$, 则
$$
\begin{aligned}
& a=m^2+m n+n^2 \\
& m^3-n^3=6-5=1 \\
& m^3-n^3=(m-n)\left(m^2+m n+n^2\right)=(m-n) a=1
\end{aligned}
$$
$\frac{1}{a}= m-n,$
$\frac{1}{a^2}= (m-n)^2=m^2-2mn+n^2$

$$
\begin{aligned}
a-\frac{1}{a^2} & =m^2+m n+n^2-\left(m^2-2 m n+n^2\right) \\
& =3 m n=3 \sqrt[3]{6} \sqrt[3]{5}=3 \sqrt[3]{30}
\end{aligned}
$$


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