Transform to axes inclined at an angle a to the original axes the equations
` x^(3) + y^(2) = r^(2)`

Answer

To transform the equation \( x^3 + y^2 = r^2 \) to axes inclined at an angle \( a \) to the original axes, we will follow these steps:

### Step 1: Define the transformation equations
We need to express the coordinates \( x \) and \( y \) in terms of the new coordinates \( x' \) and \( y' \). The transformation equations for a rotation by an angle \( a \) are:
\[ x = x' \cos a - y' \sin a \]
\[ y = x' \sin a + y' \cos a \]

### Step 2: Substitute the transformation into the original equation
We will substitute the expressions for \( x \) and \( y \) into the original equation \( x^3 + y^2 = r^2 \):
\[ (x' \cos a - y' \sin a)^3 + (x' \sin a + y' \cos a)^2 = r^2 \]

### Step 3: Expand the terms
We will expand the left-hand side of the equation:
1. For \( (x' \cos a - y' \sin a)^3 \):
\[ = (x' \cos a)^3 - 3(x' \cos a)^2(y' \sin a) + 3(x' \cos a)(y' \sin a)^2 - (y' \sin a)^3 \]
This simplifies to:
\[ = x'^3 \cos^3 a - 3x'^2 y' \cos^2 a \sin a + 3x' y'^2 \cos a \sin^2 a - y'^3 \sin^3 a \]

2. For \( (x' \sin a + y' \cos a)^2 \):
\[ = (x' \sin a)^2 + 2(x' \sin a)(y' \cos a) + (y' \cos a)^2 \]
This simplifies to:
\[ = x'^2 \sin^2 a + 2x' y' \sin a \cos a + y'^2 \cos^2 a \]

### Step 4: Combine the expanded terms
Now we combine the two expanded expressions:
\[ x'^3 \cos^3 a - 3x'^2 y' \cos^2 a \sin a + 3x' y'^2 \cos a \sin^2 a - y'^3 \sin^3 a + x'^2 \sin^2 a + 2x' y' \sin a \cos a + y'^2 \cos^2 a = r^2 \]

### Step 5: Rearranging the equation
Now we can rearrange and group the terms according to powers of \( x' \) and \( y' \). This will give us the transformed equation in terms of \( x' \) and \( y' \).

### Final Equation
The final transformed equation will be a polynomial in \( x' \) and \( y' \) that represents the same curve in the new coordinate system.