If a + b + c = 9 and ab + bc + ca = 18, then the value of
`a^(3)+b^(3)+c^(3)`+ 3abc is :

Options

A. 243
B. 254
C. 234
D. 244

Answer

To solve the problem, we need to find the value of \( a^3 + b^3 + c^3 + 3abc \) given the equations \( a + b + c = 9 \) and \( ab + bc + ca = 18 \).

### Step 1: Use the identity for the sum of cubes
We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]
From this, we can rearrange it to find:
\[ a^3 + b^3 + c^3 + 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) + 3abc \]

### Step 2: Calculate \( a^2 + b^2 + c^2 \)
We know \( a + b + c = 9 \). We can find \( a^2 + b^2 + c^2 \) using the formula:
\[ a^2 + b^2 + c^2 = (a + b + c)^2 - 2(ab + ac + bc) \]
Substituting the known values:
\[ a^2 + b^2 + c^2 = 9^2 - 2 \cdot 18 = 81 - 36 = 45 \]

### Step 3: Substitute into the identity
Now we substitute \( a + b + c \) and \( a^2 + b^2 + c^2 \) into the rearranged identity:
\[ a^3 + b^3 + c^3 + 3abc = (9)(45 - 18) + 3abc \]
Calculating \( 45 - 18 \):
\[ 45 - 18 = 27 \]
Thus,
\[ a^3 + b^3 + c^3 + 3abc = 9 \cdot 27 + 3abc = 243 + 3abc \]

### Step 4: Find \( abc \)
To find \( abc \), we can assume \( c = 0 \) (as indicated in the video transcript). Then we have:
\[ a + b = 9 \quad \text{and} \quad ab = 18 \]
Let \( a \) and \( b \) be the roots of the quadratic equation \( x^2 - (a+b)x + ab = 0 \):
\[ x^2 - 9x + 18 = 0 \]
Using the quadratic formula:
\[ x = \frac{9 \pm \sqrt{9^2 - 4 \cdot 1 \cdot 18}}{2 \cdot 1} = \frac{9 \pm \sqrt{81 - 72}}{2} = \frac{9 \pm 3}{2} \]
Thus, the roots are:
\[ x = \frac{12}{2} = 6 \quad \text{and} \quad x = \frac{6}{2} = 3 \]
So, \( a = 6 \) and \( b = 3 \) (or vice versa), and since \( c = 0 \), we have \( abc = 6 \cdot 3 \cdot 0 = 0 \).

### Step 5: Final calculation
Now substituting \( abc = 0 \) back into our equation:
\[ a^3 + b^3 + c^3 + 3abc = 243 + 3 \cdot 0 = 243 \]

Thus, the final answer is:
\[ \boxed{243} \]