Solve the following equation:

$$
\sqrt[3]{x+a}-\sqrt[3]{x-a}=b
$$
$$
\sqrt[3]{x+a}-\sqrt[3]{x-a}=b
$$

# First solution.

By cubing both sides of the equation, we get:

$$
\begin{gathered}
x+a-3 \sqrt[3]{(x+a)^{2}} \sqrt[3]{x-a}+3 \sqrt[3]{x+a} \sqrt[3]{(x-a)^{2}}-x+a=b^{3} \\
3 \sqrt[3]{(x+a)\left(x^{2}-a^{2}\right)}-3 \sqrt[3]{(x-a)\left(x^{2}-a^{2}\right)}=2 a-b^{3} \\
3 \sqrt[3]{x^{2}-a^{2}}[\sqrt[3]{x+a}-\sqrt[3]{x-a}]=2 a-b^{3}
\end{gathered}
$$

The expression in parentheses is equal to \( b \) according to (1), and thus:

$$
3 b \sqrt[3]{x^{2}-a^{2}}=2 a-b^{3}
$$

Cubing both sides of this equation:

$$
27 b^{3}\left(x^{2}-a^{2}\right)=\left(2 a-b^{3}\right)^{3}
$$

from which we get:

$$
x^{2}=\frac{\left(2 a-b^{3}\right)^{3}}{27 b^{3}}+a^{2}
$$

(Visnya Aladár, real school, VIII. grade, Pécs.)

Note. The expression under (3) can also be written as:

$$
x^{2}=\frac{8 a^{3}+15 a^{2} b^{3}+6 a b^{6}-b^{9}}{27 b^{3}}
$$

The numerator is divisible by \( a+b^{3} \) because substituting \( -b^{3} \) for \( a \) makes the numerator zero; thus, we get:

$$
8 a^{3}+15 a^{2} b^{3}+6 a b^{6}-b^{9}=\left(a+b^{3}\right)\left(8 a^{2}+7 a b^{3}-b^{6}\right)
$$

and since

$$
8 a^{2}+7 a b^{3}-b^{6}=\left(a+b^{3}\right)\left(8 a-b^{3}\right)
$$

we get:

$$
\begin{aligned}
& x^{2}=\frac{\left(a+b^{3}\right)^{2}\left(8 a-b^{3}\right)}{27 b^{3}} \\
& x= \pm\left(a+b^{3}\right) \sqrt{\frac{8 a-b^{3}}{27 b^{3}}}
\end{aligned}
$$

## Second solution.

The given equation can also be written as:

$$
\sqrt[3]{x+a}=b+\sqrt[3]{x-a}
$$

Cubing both sides of this equation and rearranging, we get:

$$
3 b(\sqrt[3]{x-a})^{2}+3 b^{2} \sqrt[3]{x-a}+b^{3}-2 a=0
$$

This equation is quadratic in \( \sqrt[3]{x-a} \), so:

$$
\begin{aligned}
& \sqrt[3]{x-a}=\frac{-3 b^{2} \pm \sqrt{24 a b-3 b^{4}}}{6 b} \\
& x=\left[-\frac{b}{2} \pm \sqrt{\frac{8 a-b^{2}}{12 b}}\right]^{3}+a
\end{aligned}
$$

(Geist Emil, real school, VII. grade, Györ.)

Note. If we cube the expression in parentheses, we easily get:

$$
x= \pm \frac{8 b^{3}+8 a}{12 b} \sqrt{\frac{8 a-b^{3}}{12 b}}
$$

or

$$
x= \pm\left(a+b^{3}\right) \sqrt{\frac{8 a-b^{3}}{27 b^{3}}}
$$

The problem was also solved by: Adonyi Dénes, high school, VII. grade, S.-A.-Ujhely, Friedmann Bernát, high school, VII. grade, S.-A.-Ujhely, Goldstein Zsigmond, high school, VII. grade, Nyíregyháza; Grünhut Béla, real school, VII. grade, Pécs; Hofbauer Ervin and Kántor Nándor, high school, VII. grade, Budapest; Messinger Ábrahám, high school, VII. grade, and Weisz Hermann, high school, VI. grade, S.-A.-Ujhely.