# 3. (10 points) Determine the sign of the expression

$\sqrt{25 \sqrt{7}-27 \sqrt{6}}-\sqrt{17 \sqrt{5}-38} \quad$ or $\quad(\sqrt{25 \sqrt{7}-28 \sqrt{6}}-\sqrt{17 \sqrt{5}-38})^{* * *}$

## Solution:

1st method: Each radicand is the cube of the difference of two numbers:

$$
\begin{gathered}
25 \sqrt{7}-27 \sqrt{6}=(\sqrt{7}-\sqrt{6})^{3} \\
17 \sqrt{5}-38=(\sqrt{5}-2)^{3}
\end{gathered}
$$

Then $\sqrt{25 \sqrt{7}-27 \sqrt{6}}-\sqrt{17 \sqrt{5}-38}=(\sqrt{7}-\sqrt{6})^{3 / 2}-(\sqrt{5}-2)^{3 / 2}$

The sign of the expression is determined by the sign of the difference of the numbers

$(\sqrt{7}-\sqrt{6})-(\sqrt{5}-2)=(\sqrt{7}+2)-(\sqrt{6}+\sqrt{5})$

Compare the numbers

$\sqrt{7}+2 \vee \sqrt{6}+\sqrt{5} \Leftrightarrow 7+4 \sqrt{7}+4 \vee 6+2 \sqrt{30}+5 \Leftrightarrow$

$$
4 \sqrt{7} \vee 2 \sqrt{30} \Leftrightarrow 112 \vee 120
$$

$112<120 \Rightarrow(\sqrt{7}-\sqrt{6})<(\sqrt{5}-2) \Rightarrow \sqrt{25 \sqrt{7}-27 \sqrt{6}}-\sqrt{17 \sqrt{5}-38}<0$
Answer: $\sqrt{25 \sqrt{7}-27 \sqrt{6}}-\sqrt{17 \sqrt{5}-38}<0$

2nd method: Compare the expressions under the square roots $25 \sqrt{7}-27 \sqrt{6} \vee 17 \sqrt{5}-38$

Since both numbers are positive, we square them and gradually eliminate the roots. As a result, we find that the minuend is greater than the subtrahend. Therefore, the sign of the difference is negative.