If a + b + c = 7 and ab +bc + ca = 1, then `a^(3) + b^(3) + c^(3) - 3abc` is equal to :

Options

A. A)322
B. B)325
C. C)412
D. D)422

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 - 3abc \) given the equations \( a + b + c = 7 \) and \( ab + bc + ca = 1 \).

We can use the identity:

\[ a^3 + b^3 + c^3 - 3abc = (a + b + c) \left( (a + b + c)^2 - 3(ab + ac + bc) \right) \]

### Step 1: Substitute the known values into the identity.

We know:
- \( a + b + c = 7 \)
- \( ab + ac + bc = 1 \)

Substituting these values into the identity:

\[ a^3 + b^3 + c^3 - 3abc = (7) \left( (7)^2 - 3(1) \right) \]

### Step 2: Calculate \( (7)^2 - 3(1) \).

First, calculate \( (7)^2 \):

\[ (7)^2 = 49 \]

Now, calculate \( 3(1) \):

\[ 3(1) = 3 \]

Now, substitute these values back into the equation:

\[ (7)^2 - 3(1) = 49 - 3 = 46 \]

### Step 3: Multiply by \( a + b + c \).

Now we substitute back into the equation:

\[ a^3 + b^3 + c^3 - 3abc = 7 \cdot 46 \]

Calculating this gives:

\[ 7 \cdot 46 = 322 \]

### Final Answer:

Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( \boxed{322} \).

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