If the zeros of the polynomial `f(x) = x^(3) - 3x^(2) - 6x +8` are of the form a-b,a,a + b, find all the zeros.

Answer

To find the zeros of the polynomial \( f(x) = x^3 - 3x^2 - 6x + 8 \) given that the zeros are of the form \( a-b, a, a+b \), we will follow these steps:

### Step 1: Define the zeros
Let the zeros be:
- \( \alpha = a - b \)
- \( \beta = a \)
- \( \gamma = a + b \)

### Step 2: Use the sum of the zeros
According to Vieta's formulas, the sum of the zeros of the polynomial \( f(x) = x^3 + bx^2 + cx + d \) is given by:
\[ \alpha + \beta + \gamma = -\frac{\text{coefficient of } x^2}{\text{coefficient of } x^3} \]
For our polynomial, this is:
\[ \alpha + \beta + \gamma = -\frac{-3}{1} = 3 \]
Substituting the values of the zeros:
\[ (a - b) + a + (a + b) = 3 \]
This simplifies to:
\[ 3a = 3 \implies a = 1 \]

### Step 3: Use the product of the zeros
The product of the zeros is given by:
\[ \alpha \cdot \beta \cdot \gamma = -\frac{\text{constant term}}{\text{coefficient of } x^3} \]
For our polynomial, this is:
\[ \alpha \cdot \beta \cdot \gamma = -\frac{8}{1} = -8 \]
Substituting the values of the zeros:
\[ (a - b) \cdot a \cdot (a + b) = -8 \]
Substituting \( a = 1 \):
\[ (1 - b) \cdot 1 \cdot (1 + b) = -8 \]
This simplifies to:
\[ (1 - b)(1 + b) = -8 \]
\[ 1 - b^2 = -8 \implies b^2 = 9 \implies b = 3 \text{ or } b = -3 \]

### Step 4: Find the zeros
Now we will find the zeros for both values of \( b \).

**Case 1: \( b = 3 \)**
- \( \alpha = 1 - 3 = -2 \)
- \( \beta = 1 \)
- \( \gamma = 1 + 3 = 4 \)

So the zeros are \( -2, 1, 4 \).

**Case 2: \( b = -3 \)**
- \( \alpha = 1 - (-3) = 4 \)
- \( \beta = 1 \)
- \( \gamma = 1 + (-3) = -2 \)

So the zeros are \( 4, 1, -2 \).

### Conclusion
In both cases, the zeros are the same: \( -2, 1, 4 \).

### Final Answer
The zeros of the polynomial \( f(x) = x^3 - 3x^2 - 6x + 8 \) are \( -2, 1, 4 \).

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