根式化简求整数a,b的值

郝酒

设整数$a,b$满足$$3\sqrt{\sqrt[3]{5}-\sqrt[3]{4}}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{2},$$求$a,b$的值.

hbghlyj

math.stackexchange.com/questions/1395879/calc … qrt32-sqrt320-sqrt32
math.stackexchange.com/questions/3492510/find … t-sqrt-sqrt35-sqrt34
math.stackexchange.com/questions/1078705/cube-roots-escape
You can check the following expression works by squaring the right hand side
$$
\sqrt{\sqrt[3]{5}-\sqrt[3]{4}} \times 3 = \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25} \, .
$$
So $ a + b + c = 47 $.

As for finding it: It is initially a good idea to try working in the field extension $ \mathbb{Q}(\sqrt[3]{4}, \sqrt[3]{5}) = \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{5}) $, which has a basis given by
$$
\{ 1, \sqrt[3]{2}, \sqrt[3]{4}, \sqrt[3]{1\times5}, \sqrt[3]{2\times5}, \sqrt[3]{4\times5}, \sqrt[3]{1\times25}, \sqrt[3]{2\times25}, \sqrt[3]{4\times25} \} \, .
$$

Then you want to consider the equation
$$
  3\sqrt{\sqrt[3]{5} - \sqrt[3]{4}} = r + s \sqrt[3]{2} + t \sqrt[3]{4} + u \sqrt[3]{5} + v \sqrt[3]{10} + w \sqrt[3]{20} + x \sqrt[3]{25} + y \sqrt[3]{50} + z \sqrt[3]{100} \, ,
$$ with $ r, \ldots, z \in \mathbb{Q} $.  In particular you want $ r, \ldots, z \in \{ -1, 0, 1 \} $ with exactly three of them non-zero.  Square both sides, and compare the coefficient of $ \sqrt[3]{5} $ and $ \sqrt[3]{4} $ on both sides.  We get
\begin{align}
\sqrt[3]{5} &: \quad 9 = 5x^2 + 2ru + 4tv + 4sw + 20yz \\
\sqrt[3]{4} &: \quad -9 = s^2 + 2rt + 10wx + 10vy + 10uz \, .
\end{align}

You've assumed that all the non-zero coefficients are $ \pm 1 $, so we must have $ x = \pm 1 $ and $ s = \pm 1 $, by parity considerations.  By assumption there are only 3 non-zero coefficients, $ s, x $ and one other.  So $ ru = tv = yz = 0 $ and $ rt = vy = uz = 0 $ and the equations reduce to
\begin{align}
\sqrt[3]{5} &: \quad 9 = 5x^2 + 4sw \\
\sqrt[3]{4} &: \quad -9 = s^2 + 10wx \, .
\end{align}

We must have $ sw = 1 $ and $ wx = -1 $, so $ s = w = -x $.  This gives us two possibilities by taking $ x = 1 $ or $ x = - 1 $.  The form of the solution you are looking for has two $ +1 $ coefficients.  This must come from $ x = -1 $, and be
$$
  \sqrt[3]{2} + \sqrt[3]{20} - \sqrt[3]{25}
$$

Now you need to go back and check this actually does work as a solution because we've made many assumptions in finding it.  In particular we've assumed that such a solution exists and lives in $ \mathbb{Q}(\sqrt[3]{2}, \sqrt[3]{5}) $, rather than some larger field extension.  And we've only used two equations which come from comparing coefficients; do the rest also work?

郝酒

多谢，学习了