3. Find all natural numbers for which

$$
(\sqrt[3]{a}+\sqrt[3]{b}-1)^{2}=49+20 \sqrt[3]{6}
$$
Solution. Let's introduce the substitution $\sqrt[3]{a}=x \sqrt[3]{6}, \sqrt[3]{b}=y \sqrt[3]{36}$.

Then

$$
\begin{aligned}
(x \sqrt[3]{6}+y \sqrt[3]{36}-1)^{2} & =x^{2} \sqrt[3]{36}+6 y^{2} \sqrt[3]{6}+1+12 x y-2 x \sqrt[3]{6}-2 y \sqrt[3]{36} \\
& =\left(x^{2}-2 y\right) \sqrt[3]{36}+\left(6 y^{2}-2 x\right) \sqrt[3]{6}+12 x y+1
\end{aligned}
$$

From

$$
\left(x^{2}-2 y\right) \sqrt[3]{36}+\left(6 y^{2}-2 x\right) \sqrt[3]{6}+12 x y+1=49+20 \sqrt[3]{6}
$$

we obtain the system

$$
\begin{aligned}
& x^{2}-2 y=0 \\
& 6 y^{2}-2 x=20 \\
& 12 x y+1=49
\end{aligned}
$$

whose solution is $x=y=2$. Thus, $a=48, b=288$, and due to symmetry, the solution is also $a=288, b=48$.