To solve the problem and find the value of \( A \), we need to use some properties of H.C.F (Highest Common Factor) and L.C.M (Least Common Multiple) of polynomials.
Given:
- \( \text{H.C.F} \) of \( x^3 - 1 \) and \( A \) is \( x - 1 \)
- \( \text{L.C.M} \) of \( x^3 - 1 \) and \( A \) is \( x^6 - 1 \)
We use the relationship between H.C.F and L.C.M of two expressions \( f(x) \) and \( g(x) \):
[tex]\[ \text{H.C.F}(f(x), g(x)) \cdot \text{L.C.M}(f(x), g(x)) = f(x) \cdot g(x) \][/tex]
Applying this to our problem:
[tex]\[ (x - 1) \cdot (x^6 - 1) = (x^3 - 1) \cdot A \][/tex]
First, we factorize the expressions where necessary.
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \][/tex]
[tex]\[ x^6 - 1 = (x^3 - 1)(x^3 + 1) = (x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1) \][/tex]
Given:
[tex]\[ (x - 1) \cdot (x^6 - 1) = (x^3 - 1) \cdot A \][/tex]
Substitute the factored forms:
[tex]\[ (x - 1) \left((x - 1)(x^2 + x + 1)(x + 1)(x^2 - x + 1)\right) = ((x - 1)(x^2 + x + 1)) \cdot A \][/tex]
[tex]\[ (x - 1)^2(x^2 + x + 1)(x + 1)(x^2 - x + 1) = (x - 1)(x^2 + x + 1) \cdot A \][/tex]
We can cancel \((x - 1)(x^2 + x + 1)\) from both sides, as they are common factors:
[tex]\[ (x - 1)(x + 1)(x^2 - x + 1) = A \][/tex]
So we need to find \( A \):
\( A = (x - 1)(x + 1)(x^2 - x + 1) \)
[tex]\[ A = (x^2 - 1)(x^2 - x + 1) \][/tex]
The given choices are:
1. \( x^3 + 1 \)
2. \( x^4 - x^3 + x - 1 \)
3. \( (x - 1)(x^2 - x + 1) \)
4. \( (x - 1)(x^2 + x + 1) \)
Let's test each choice to match our solution for \( A \).
By comparing our final expression of A with the provided options, it should be clear:
Given \( A = (x - 1)(x^2 - x + 1) \).
Therefore, the answer is option 3:
[tex]\[ (x - 1)(x^2 - x + 1) \][/tex]
The correct value of \( A \) is:
[tex]\[ \boxed{(x-1)(x^2 - x + 1)} \][/tex]
