89. Simplify the expression:

a) $\sqrt{|40 \sqrt{2}-57|}-\sqrt{40 \sqrt{2}+57}$

b) $\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}$
Solution. a) Let $\sqrt{57-40 \sqrt{2}}-\sqrt{57+40 \sqrt{\overline{2}}}=x ; x<0$, since $\sqrt{57-40 \sqrt{2}}<\sqrt{57+40 \sqrt{2}}$.

Square both sides of the equation:

$$
x^{2}=57-40 \sqrt{2}-2 \sqrt{57^{2}-(40 \sqrt{2})^{2}}+57+40 \sqrt{2}
$$

from which

$$
x^{2}=114-2 \sqrt{49}, x^{2}=100, x=-10
$$

(Solution by the method of undetermined coefficients see p. 51.)

b) Let the value of the expression be $x$, i.e.

$$
\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}}=x
$$

Cube both sides of the equation:

$$
\begin{aligned}
& 4+3 \cdot(\sqrt[3]{2+\sqrt{5}})^{2} \cdot \sqrt[3]{2-\sqrt{5}}+ \\
& +3 \cdot \sqrt[3]{2+\sqrt{5}} \cdot(\sqrt[3]{2-\sqrt{5}})^{2}=x^{3}
\end{aligned}
$$

or

$$
3 \sqrt[3]{2+\sqrt{5}} \cdot \sqrt[3]{2-\sqrt{5}} \cdot(\sqrt[3]{2+\sqrt{5}}+\sqrt[3]{2-\sqrt{5}})=x^{3}-4
$$

or (see (1))

$$
\begin{aligned}
3 \cdot \sqrt[3]{2^{2}-(\sqrt{5})^{2}} \cdot x & =x^{3}-4, \text { i.e. } \\
x^{3}+3 x-4 & =0
\end{aligned}
$$

Thus, the value $x$ of the expression is a root of equation (2).

It is obvious that the number $x=1$ is a root of equation (2).