If a = 29, b = 24 and c = 27, find the value of `a^3+b^3+c^3-3abc`

Options

A. 1620
B. 1640
C. 1520
D. 1580

Answer

To find the value of \( a^3 + b^3 + c^3 - 3abc \) given \( a = 29 \), \( b = 24 \), and \( c = 27 \), we can use the identity:

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

### Step 1: Calculate \( a + b + c \)

\[ a + b + c = 29 + 24 + 27 \]

\[ = 80 \]

### Step 2: Calculate \( a - b \), \( b - c \), and \( c - a \)

\[ a - b = 29 - 24 = 5 \]

\[ b - c = 24 - 27 = -3 \]

\[ c - a = 27 - 29 = -2 \]

### Step 3: Calculate \( (a - b)^2 \), \( (b - c)^2 \), and \( (c - a)^2 \)

\[ (a - b)^2 = 5^2 = 25 \]

\[ (b - c)^2 = (-3)^2 = 9 \]

\[ (c - a)^2 = (-2)^2 = 4 \]

### Step 4: Sum the squares

\[ (a - b)^2 + (b - c)^2 + (c - a)^2 = 25 + 9 + 4 \]

\[ = 38 \]

### Step 5: Substitute back into the identity

Now substitute \( a + b + c \) and the sum of squares into the identity:

\[ a^3 + b^3 + c^3 - 3abc = \frac{1}{2} (80) (38) \]

### Step 6: Calculate the final value

\[ = 40 \times 38 \]

\[ = 1520 \]

Thus, the value of \( a^3 + b^3 + c^3 - 3abc \) is \( 1520 \).