The value of `a^3+b^3+c^3-3abc`, when a =87, b = -126 and c=39 is :

Options

A. 0
B. 1259
C. -48
D. none of these

Explanation

To find the value of \( a^3 + b^3 + c^3 - 3abc \) for the given values \( a = 87 \), \( b = -126 \), and \( c = 39 \), 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 1: Calculate \( a + b + c \)

\[ a + b + c = 87 + (-126) + 39 \]

Calculating this step-by-step:

1. \( 87 - 126 = -39 \)
2. \( -39 + 39 = 0 \)

So,

\[ a + b + c = 0 \]

### Step 2: Use the identity

Since \( a + b + c = 0 \), we can substitute this into the identity:

\[ a^3 + b^3 + c^3 - 3abc = (0)(a^2 + b^2 + c^2 - ab - ac - bc) \]

### Step 3: Simplify the expression

Since any number multiplied by zero is zero, we have:

\[ a^3 + b^3 + c^3 - 3abc = 0 \]

### Final Answer

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

\[ \boxed{0} \]