If bc `(b+c) + ca (c +a) +ab (a+b) + a ^(3) + b ^(3) + c ^(3)` divided by `a ^(2) + b ^(2) + c ^(2),` then the quotient is :

Options

A. `a +b +c`
B. `a ^(2) +b ^(2) + c ^(2)`
C. `sqrt (a ^(2) +b ^(2) + c^(2))`
D. `a ^(3) + b ^(3) + c ^(3)`

Answer

To solve the problem, we need to simplify the expression given in the question and then divide it by \( a^2 + b^2 + c^2 \).

### Step-by-Step Solution:

1. **Write the Expression**: We start with the expression:
\[ bc(b+c) + ca(c+a) + ab(a+b) + a^3 + b^3 + c^3 \]

2. **Expand Each Term**:
- For \( bc(b+c) \):
\[ bc(b+c) = b^2c + bc^2 \]
- For \( ca(c+a) \):
\[ ca(c+a) = c^2a + ca^2 \]
- For \( ab(a+b) \):
\[ ab(a+b) = a^2b + ab^2 \]
- The terms \( a^3, b^3, c^3 \) remain as they are.

Now, combining all these expanded terms:
\[ b^2c + bc^2 + c^2a + ca^2 + a^2b + ab^2 + a^3 + b^3 + c^3 \]

3. **Group the Terms**: We can group the terms based on their degrees:
\[ a^3 + b^3 + c^3 + (a^2b + ab^2) + (b^2c + bc^2) + (c^2a + ca^2) \]

4. **Factor Each Group**:
- The terms \( a^3 + b^3 + c^3 \) can be factored using the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
- The terms \( a^2b + ab^2 \) can be factored as:
\[ ab(a + b) \]
- The terms \( b^2c + bc^2 \) can be factored as:
\[ bc(b + c) \]
- The terms \( c^2a + ca^2 \) can be factored as:
\[ ca(c + a) \]

5. **Combine the Factored Terms**:
The entire expression can be rewritten as:
\[ (a+b+c)(a^2 + b^2 + c^2) + ab(a+b) + bc(b+c) + ca(c+a) \]
However, we can see that the sum of products \( ab(a+b) + bc(b+c) + ca(c+a) \) can be rearranged to fit into the form of \( a^2 + b^2 + c^2 \).

6. **Final Expression**:
The expression simplifies to:
\[ (a+b+c)(a^2 + b^2 + c^2) \]

7. **Divide by \( a^2 + b^2 + c^2 \)**:
Now, we divide the entire expression by \( a^2 + b^2 + c^2 \):
\[ \frac{(a+b+c)(a^2 + b^2 + c^2)}{a^2 + b^2 + c^2} = a + b + c \]

### Conclusion:
The quotient is:
\[ \boxed{a + b + c} \]