If `x^a xx x^b xx x^c = 1`, then `a^3 + b^3 + c^3` is equal to :

Options

A. 9
B. abc
C. `a + b + c`
D. `3abc`

Answer

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

1. **Given Equation**:
\[ x^a \cdot x^b \cdot x^c = 1 \]

2. **Combine Exponents**:
Since the bases are the same, we can add the exponents:
\[ x^{a + b + c} = 1 \]

3. **Equate to Zero**:
We know that \( x^0 = 1 \) for any non-zero \( x \). Therefore, we can set the exponent equal to zero:
\[ a + b + c = 0 \]

4. **Use the Identity for Cubes**:
We need to find \( a^3 + b^3 + c^3 \). We can use the identity:
\[ a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - ac - bc) \]

5. **Substitute \( a + b + c \)**:
Since \( a + b + c = 0 \), we substitute this into the identity:
\[ a^3 + b^3 + c^3 - 3abc = 0 \cdot (a^2 + b^2 + c^2 - ab - ac - bc) \]
This simplifies to:
\[ a^3 + b^3 + c^3 - 3abc = 0 \]

6. **Solve for \( a^3 + b^3 + c^3 \)**:
Rearranging gives us:
\[ a^3 + b^3 + c^3 = 3abc \]

Thus, the final result is:
\[ a^3 + b^3 + c^3 = 3abc \]