If (a+c+1)=0, then find the value of `(a^3+c^3+1-3ac)`
A. `-1`
B. 1
C. 2
D. 0

Options

A. A
B. C
C. B
D. D

Answer

To solve the problem, we start with the given equation:

1. **Given Equation**:
\[ a + c + 1 = 0 \]
From this, we can express \(a + c\) as:
\[ a + c = -1 \]

2. **Using the Identity for Cubes**:
We know the identity for the sum of cubes:
\[ a^3 + c^3 = (a + c)(a^2 - ac + c^2) \]
We can also use the fact that:
\[ a^2 + c^2 = (a + c)^2 - 2ac \]
Substituting \(a + c = -1\):
\[ a^2 + c^2 = (-1)^2 - 2ac = 1 - 2ac \]

3. **Substituting into the Cube Identity**:
Now, substituting \(a + c\) into the identity:
\[ a^3 + c^3 = (-1)(a^2 - ac + c^2) \]
We can replace \(a^2 + c^2\) in the equation:
\[ a^3 + c^3 = -1 \left( (1 - 2ac) - ac \right) \]
Simplifying this gives:
\[ a^3 + c^3 = -1(1 - 3ac) = -1 + 3ac \]

4. **Finding the Required Expression**:
We need to find the value of:
\[ a^3 + c^3 + 1 - 3ac \]
Substituting \(a^3 + c^3\) from above:
\[ a^3 + c^3 + 1 - 3ac = (-1 + 3ac) + 1 - 3ac \]
This simplifies to:
\[ 0 \]

5. **Final Answer**:
Therefore, the value of \(a^3 + c^3 + 1 - 3ac\) is:
\[ \boxed{0} \]