To solve the problem, we need to find the relationship between the roots \( p, q, r \) of the cubic equation \( x^3 + 2x^2 + 3x + 3 = 0 \).
### Step-by-step Solution:
1. **Identify the roots of the cubic equation**:
The given cubic equation is:
\[
x^3 + 2x^2 + 3x + 3 = 0
\]
Let \( p, q, r \) be the roots of this equation.
2. **Use Vieta's formulas**:
According to Vieta's formulas, for a cubic equation \( ax^3 + bx^2 + cx + d = 0 \):
- The sum of the roots \( p + q + r = -\frac{b}{a} \)
- The sum of the products of the roots taken two at a time \( pq + qr + rp = \frac{c}{a} \)
- The product of the roots \( pqr = -\frac{d}{a} \)
For our equation:
- \( a = 1, b = 2, c = 3, d = 3 \)
Thus, we have:
- \( p + q + r = -\frac{2}{1} = -2 \)
- \( pq + qr + rp = \frac{3}{1} = 3 \)
- \( pqr = -\frac{3}{1} = -3 \)
3. **Express the roots in terms of a new variable**:
We define:
\[
y_1 = \frac{p}{p + 1}, \quad y_2 = \frac{q}{q + 1}, \quad y_3 = \frac{r}{r + 1}
\]
4. **Relate the new variables to the original roots**:
From the definitions above, we can express \( p, q, r \) in terms of \( y_1, y_2, y_3 \):
\[
p = \frac{y_1}{1 - y_1}, \quad q = \frac{y_2}{1 - y_2}, \quad r = \frac{y_3}{1 - y_3}
\]
5. **Substitute back into the cubic equation**:
Since \( p, q, r \) are roots of the cubic equation, we substitute these expressions into the equation and simplify.
6. **Find the sum of the new variables**:
We need to find the relationship:
\[
y_1 + y_2 + y_3 = 5
\]
This can be derived from the earlier steps and using Vieta's relations.
7. **Conclusion**:
The final relationship we find is:
\[
\frac{p}{p + 1} + \frac{q}{q + 1} + \frac{r}{r + 1} = 5
\]
