If `a + b+c = 4` and `ab + bc + ca = 2`, then `a^(3) + b^(3) + c^(3) - 3abc` is equal to :

Options

A. 36
B. 32
C. 48
D. 40

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given that \( a + b + c = 4 \) and \( ab + bc + ca = 2 \).

### Step 1: Use the identity for \( a^3 + b^3 + c^3 - 3abc \)

We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 2: Calculate \( a^2 + b^2 + c^2 \)

We know that:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the known values:
\[ a + b + c = 4 \quad \text{and} \quad ab + ac + bc = 2 \]
We calculate:
\[ a^2 + b^2 + c^2 = 4^2 - 2 \cdot 2 = 16 - 4 = 12 \]

### Step 3: Substitute into the identity

Now we substitute \( a + b + c \) and \( a^2 + b^2 + c^2 \) into the identity:
\[ a^3 + b^3 + c^3 - 3abc = (4)(12 - 2) = 4 \cdot 10 = 40 \]

### Conclusion

Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( 40 \).