If `a = 1, b = 2 and c =-3`, then the value of
`(a ^(3) + b ^(3) + c ^(3) - 3 abc)/( ab + bc + ca - (a ^(2) + b ^(2) + c ^(2)))` is :

Options

A. 3
B. 2
C. 0
D. 1

Answer

To solve the expression

\[ \frac{a^3 + b^3 + c^3 - 3abc}{ab + bc + ca - (a^2 + b^2 + c^2)} \]

given \(a = 1\), \(b = 2\), and \(c = -3\), we will follow these steps:

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

First, we calculate the sum of \(a\), \(b\), and \(c\):

\[ a + b + c = 1 + 2 - 3 = 0 \]

**Hint:** If the sum of \(a\), \(b\), and \(c\) is zero, we can use the identity \(a^3 + b^3 + c^3 = 3abc\).

### Step 2: Calculate \(a^3 + b^3 + c^3\)

Using the identity mentioned, we find:

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

Calculating \(abc\):

\[ abc = 1 \cdot 2 \cdot (-3) = -6 \]

Thus,

\[ a^3 + b^3 + c^3 = 3 \cdot (-6) = -18 \]

**Hint:** Remember that \(a^3 + b^3 + c^3\) can be simplified when \(a + b + c = 0\).

### Step 3: Calculate \(ab + bc + ca\)

Now, we calculate \(ab + bc + ca\):

\[ ab = 1 \cdot 2 = 2 \]
\[ bc = 2 \cdot (-3) = -6 \]
\[ ca = (-3) \cdot 1 = -3 \]

Adding these together:

\[ ab + bc + ca = 2 - 6 - 3 = -7 \]

**Hint:** Multiply the pairs of \(a\), \(b\), and \(c\) to find \(ab\), \(bc\), and \(ca\).

### Step 4: Calculate \(a^2 + b^2 + c^2\)

Next, we calculate \(a^2 + b^2 + c^2\):

\[ a^2 = 1^2 = 1 \]
\[ b^2 = 2^2 = 4 \]
\[ c^2 = (-3)^2 = 9 \]

Adding these together:

\[ a^2 + b^2 + c^2 = 1 + 4 + 9 = 14 \]

**Hint:** Square each value of \(a\), \(b\), and \(c\) and then sum them.

### Step 5: Substitute into the expression

Now we substitute back into the original expression:

\[ \frac{a^3 + b^3 + c^3 - 3abc}{ab + bc + ca - (a^2 + b^2 + c^2)} = \frac{-18 - 3(-6)}{-7 - 14} \]

Calculating the numerator:

\[ -18 + 18 = 0 \]

Calculating the denominator:

\[ -7 - 14 = -21 \]

So we have:

\[ \frac{0}{-21} = 0 \]

### Final Answer

The value of the expression is

\[ \boxed{0} \]