Solve the following system of equations:

$$
(x+y)\left(x^{2}+y^{2}\right)=a,
$$

$$
(x-y)\left(x^{2}-y^{2}\right)=b,
$$
I. Solution: By transforming the left sides into polynomials,

$$
\begin{aligned}
& x^{3}+x^{2} y+x y^{2}+y^{3}=a \\
& x^{3}-x^{2} y-x y^{2}+y^{3}=b
\end{aligned}
$$

we arrive at these equations. Since both equations are cubic and neither appears to be reducible to a lower degree, we will need to take cube roots when solving the system of equations. Therefore, we aim to express the variables in the simplest form possible, forming a complete cube.

By adding (1) and (2) and then dividing by two,

$$
x^{3}+y^{3}=\frac{a+b}{2}
$$

The left side would be completed to the cube of $(x+y)$ by $3 x^{2} y+3 x y^{2}$. Notice that if we subtract (2), we arrive at an expression of exactly this form:

$$
2 x^{2} y+2 x y^{2}=a-b
$$

Multiplying both sides by $\frac{3}{2}$, we get

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

The sum of (3) and (4) is

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

Taking the cube root,

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

On the other hand, (4) can be transformed as follows:

$$
x y(x+y)=\frac{a-b}{2}
$$

Substituting the value of $x+y$ obtained from (5),

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

Thus, $x$ and $y$ are the roots of the following quadratic equation

$$
z^{2}-\sqrt[3]{2 a-b} \cdot z+\frac{a-b}{2 \sqrt[3]{2 a-b}}=0
$$

Solving the equation:

$$
z_{1}=x_{1}=y_{2}=\frac{\sqrt[3]{2 a-b}+\sqrt{\sqrt[3]{(2 a-b)^{2}}}-\frac{2(a-b)}{\sqrt[3]{2 a-b}}}{2}, \sqrt[\sqrt[3]{2 a-b}-\sqrt{\sqrt[3]{(2 a-b)^{2}}-\frac{2(a-b)}{\sqrt[3]{2 a-b}}}]{2}
$$

II. Solution: Introduce new variables. Let

$$
x+y=u, \quad \text { and } \quad x y=v
$$

For the purpose of substituting the new variables, we transform our equations

$$
\begin{gathered}
(x+y)\left(x^{2}+y^{2}\right)=(x+y)\left[(x+y)^{2}-2 x y\right]=a \\
(x-y)\left(x^{2}-y^{2}\right)=(x-y)^{2}(x+y)=\left[(x+y)^{2}-4 x y\right](x+y)=b
\end{gathered}
$$

Performing the substitution,

$$
\begin{aligned}
& u\left(u^{2}-2 v\right)=u^{3}-2 u v=a \\
& \left(u^{2}-4 v\right) u=u^{3}-4 u v=b
\end{aligned}
$$

Subtract (2) from twice (1)

$$
u^{3}=2 a-b ; \quad u=\sqrt[3]{2 a-b}=x+y
$$

Express $v$ in terms of $u$ from (1), then substitute the value of $u$ from (3)

$$
v=\frac{u^{3}-a}{2 u}=\frac{a-b}{2 \sqrt[3]{2 a-b}}=x y
$$

Since (3) and (4) are identical to equations (5) and (6) of the first solution, we can proceed with the calculations as in the first solution.

Note: Both given functions are symmetric expressions of $x$ and $y$, meaning they remain unchanged when $x$ and $y$ are swapped. The expressions $u=x+y$ and $v=x y$ are called the elementary symmetric expressions of the two variables. It can be proven that any symmetric expression can be uniquely expressed using the elementary symmetric expressions with just the four basic arithmetic operations. This is what we did in the second solution.