If `a+b+c=6` and `a^2+b^2+c^2=38`, then whatis the value of `a(b^2+c^(2))+b(c^2+a^2)+c(a^2+b^2)+3abc?`

Options

A. 3
B. `-3`
C. 6
D. `-6`

Answer

To solve the problem step by step, we start with the given equations:

1. **Given Equations**:
- \( a + b + c = 6 \) (Equation 1)
- \( a^2 + b^2 + c^2 = 38 \) (Equation 2)

2. **Using the Identity**:
We can use the identity:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + bc + ca) \]
Substituting the values from Equation 1 into the identity:
\[ 38 = 6^2 - 2(ab + bc + ca) \]
Simplifying this gives:
\[ 38 = 36 - 2(ab + bc + ca) \]
Rearranging gives:
\[ 2(ab + bc + ca) = 36 - 38 = -2 \]
Therefore:
\[ ab + bc + ca = -1 \quad (Equation 3) \]

3. **Finding the Expression**:
We need to find the value of:
\[ a(b^2 + c^2) + b(c^2 + a^2) + c(a^2 + b^2) + 3abc \]
We can rewrite \( b^2 + c^2 \) using the identity:
\[ b^2 + c^2 = (b + c)^2 - 2bc \]
Since \( b + c = 6 - a \), we have:
\[ b^2 + c^2 = (6 - a)^2 - 2bc \]
Similarly, we can express \( c^2 + a^2 \) and \( a^2 + b^2 \) in terms of \( a, b, c \).

4. **Substituting Values**:
We can substitute \( b + c \) and \( a + c \) and \( a + b \) into the expression:
\[ a((6 - a)^2 - 2bc) + b((6 - b)^2 - 2ac) + c((6 - c)^2 - 2ab) + 3abc \]
This expression can be simplified further, but we can also directly calculate the value using the known values of \( ab + bc + ca \) and \( a + b + c \).

5. **Final Calculation**:
We can now substitute \( ab + bc + ca = -1 \) into the expression:
\[ a(b^2 + c^2) + b(c^2 + a^2) + c(a^2 + b^2) + 3abc = (a + b + c)(a^2 + b^2 + c^2) - 3abc \]
Substituting the known values:
\[ = 6 \cdot 38 - 3abc \]
Now we need to find \( abc \). Since we have \( ab + bc + ca = -1 \), we can assume \( c = 0 \) for simplicity, leading to \( a + b = 6 \) and \( a^2 + b^2 = 38 \). Solving these gives \( ab = -1 \).

6. **Final Value**:
Thus, substituting back gives:
\[ = 228 - 3(-1) = 228 + 3 = 231 \]

Thus, the final answer is:
\[ \boxed{231} \]