Solve the following system of equations:

$$
\begin{aligned}
& a \sqrt{a}+b \sqrt{b}=183 \\
& a \sqrt{b}+b \sqrt{a}=182
\end{aligned}
$$
I. solution. Let $x=\sqrt{a}$ and $y=\sqrt{b}$, then the system of equations becomes:

$$
\begin{aligned}
x^{3}+y^{3} & =183 \\
x^{2} y+y^{2} x & =182
\end{aligned}
$$

Using the fact that

$$
(x+y)^{3}=\left(x^{3}+y^{3}\right)+3\left(x^{2} y+y^{2} x\right)=183+3 \cdot 182=729=9^{3}
$$

we get that $x+y=9$. Writing the difference of (1) and (2):

$$
1=(x+y)\left(x^{2}-x y+y^{2}\right)-x y(x+y)=(x+y)\left(x^{2}-2 x y+y^{2}\right)=(x+y)(x-y)^{2}
$$

Substituting the previously obtained value of $x+y$, we get that $(x-y)^{2}=\frac{1}{9}$. There are two cases:

Case I: $x+y=9$ and $x-y=\frac{1}{3}$. From this, $x=\frac{14}{3}$ and $y=\frac{13}{3}$, hence $a=\frac{196}{9}$ and $b=\frac{169}{9}$.

Case II: $x+y=9$ and $x-y=-\frac{1}{3}$. Then $x=\frac{13}{3}$ and $y=\frac{14}{3}$, from which $a=\frac{169}{9}$ and $b=\frac{196}{9}$.

Remark: We can also reach the goal by expressing one of the unknowns from the equation $x+y=9$ obtained in the first step, and substituting it back into, say, equation (1).

II. solution. Subtracting 183 times the second equation from 182 times the first equation, we get:

$$
182 a \sqrt{a}+182 b \sqrt{b}-183 a \sqrt{b}-183 b \sqrt{a}=0
$$

Since $b=0$ does not provide a solution, we can divide the equation by $(\sqrt{b})^{3}$:

$$
182 \cdot \frac{(\sqrt{a})^{3}}{(\sqrt{b})^{3}}+182-183 \cdot \frac{(\sqrt{a})^{2}}{(\sqrt{b})^{2}}-183 \cdot \frac{\sqrt{a}}{\sqrt{b}}=0
$$

Introducing the notation $x=\frac{\sqrt{a}}{\sqrt{b}}$, the equation can be written as:

$$
182 x^{3}+182-183 x^{2}-183 x=0
$$

Factoring the left side, we get:

$$
(x+1)\left(182 x^{2}-365 x+182\right)=0
$$

Since $x=-1$ does not provide a solution, the two solutions are given by the roots of the quadratic equation from the second factor: $x_{1}=\frac{14}{13}$ and $x_{2}=\frac{13}{14}$.

In the first case, $\frac{\sqrt{a}}{\sqrt{b}}=\frac{14}{13}$, so $\sqrt{a}=\frac{14}{13} \sqrt{b}$, and hence $a=\frac{196}{169} b$. Substituting this into the original first equation, we get:

$$
\begin{array}{r}
\frac{196}{169} b \cdot \frac{14}{13} \sqrt{b}+b \sqrt{b}=183 \\
b \sqrt{b} \cdot\left(\frac{2744}{2197}+1\right)=b \sqrt{b} \cdot \frac{4941}{2197}=183 \\
b \sqrt{b}=\frac{402051}{4941}=\frac{2197}{27}=\left(\frac{13}{3}\right)^{3}
\end{array}
$$

Thus, $\sqrt{b}=\frac{13}{3}$, so $b=\frac{169}{9}$, and hence

$$
a=\frac{196}{169} \cdot \frac{169}{9}=\frac{196}{9}
$$

Since the original system of equations is symmetric in $a$ and $b$, in the other case we get that $a=\frac{169}{9}$ and $b=\frac{196}{9}$.