If the zeroes of the polynomial `x^3-2x^2+x+1` are a-b ,a and a+b then find the values of a and b

Answer

To find the values of \( a \) and \( b \) given that the zeros of the polynomial \( x^3 - 2x^2 + x + 1 \) are \( a-b \), \( a \), and \( a+b \), we can follow these steps:

### Step 1: Identify the polynomial and its coefficients
The polynomial is given as:
\[ P(x) = x^3 - 2x^2 + x + 1 \]
From this polynomial, we can identify the coefficients:
- \( a = 1 \) (coefficient of \( x^3 \))
- \( b = -2 \) (coefficient of \( x^2 \))
- \( c = 1 \) (coefficient of \( x \))
- \( d = 1 \) (constant term)

### Step 2: Use the relationship for the sum of the zeros
The sum of the zeros of a cubic polynomial \( ax^3 + bx^2 + cx + d \) is given by the formula:
\[ \text{Sum of zeros} = -\frac{b}{a} \]
For our polynomial:
\[ \text{Sum of zeros} = -\frac{-2}{1} = 2 \]
The zeros are \( a-b \), \( a \), and \( a+b \). Therefore, we can write:
\[ (a-b) + a + (a+b) = 2 \]
Simplifying this:
\[ 3a = 2 \]
Thus, we find:
\[ a = \frac{2}{3} \]

### Step 3: Use the relationship for the product of the zeros
The product of the zeros of a cubic polynomial is given by:
\[ \text{Product of zeros} = -\frac{d}{a} \]
For our polynomial:
\[ \text{Product of zeros} = -\frac{1}{1} = -1 \]
The product of the zeros can also be expressed as:
\[ (a-b) \cdot a \cdot (a+b) = -1 \]
Using the identity \( (a-b)(a+b) = a^2 - b^2 \), we can rewrite the product:
\[ a(a^2 - b^2) = -1 \]
Substituting \( a = \frac{2}{3} \):
\[ \frac{2}{3} \left( \left(\frac{2}{3}\right)^2 - b^2 \right) = -1 \]
Calculating \( \left(\frac{2}{3}\right)^2 \):
\[ \frac{2}{3} \left( \frac{4}{9} - b^2 \right) = -1 \]
Multiplying through by 3 to eliminate the fraction:
\[ 2 \left( \frac{4}{9} - b^2 \right) = -3 \]
Expanding:
\[ \frac{8}{9} - 2b^2 = -3 \]
Rearranging gives:
\[ 2b^2 = \frac{8}{9} + 3 \]
Converting 3 to a fraction:
\[ 3 = \frac{27}{9} \]
So:
\[ 2b^2 = \frac{8}{9} + \frac{27}{9} = \frac{35}{9} \]
Dividing by 2:
\[ b^2 = \frac{35}{18} \]

### Step 4: Solve for \( b \)
Taking the square root of both sides:
\[ b = \pm \sqrt{\frac{35}{18}} = \pm \frac{\sqrt{35}}{3\sqrt{2}} = \pm \frac{\sqrt{35}}{3\sqrt{2}} = \pm \frac{\sqrt{70}}{6} \]

### Final Values
Thus, the values of \( a \) and \( b \) are:
\[ a = \frac{2}{3}, \quad b = \pm \frac{\sqrt{70}}{6} \]