If `alpha,beta` gamma are the zeroes of polynomial f(x)`=(x-1)(x^(2)+x+3)`, then find the value of `alpha^(3)+beta^(3)+gamma^(3)`.

Explanation

To find the value of \( \alpha^3 + \beta^3 + \gamma^3 \) where \( \alpha, \beta, \gamma \) are the zeroes of the polynomial \( f(x) = (x-1)(x^2 + x + 3) \), we can follow these steps:

### Step 1: Expand the Polynomial
We start by expanding the polynomial \( f(x) \):

\[ f(x) = (x-1)(x^2 + x + 3) \]

Using the distributive property (FOIL method):

\[ f(x) = x(x^2 + x + 3) - 1(x^2 + x + 3) \]
\[ = x^3 + x^2 + 3x - (x^2 + x + 3) \]
\[ = x^3 + x^2 + 3x - x^2 - x - 3 \]
\[ = x^3 + (1 - 1)x^2 + (3 - 1)x - 3 \]
\[ = x^3 + 2x - 3 \]

### Step 2: Identify Coefficients
From the polynomial \( f(x) = x^3 + 2x - 3 \), we can identify the coefficients:
- \( a = 1 \) (coefficient of \( x^3 \))
- \( b = 0 \) (coefficient of \( x^2 \))
- \( c = 2 \) (coefficient of \( x \))
- \( d = -3 \) (constant term)

### Step 3: Use the Relationships of Roots
Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta + \gamma = -\frac{b}{a} = -\frac{0}{1} = 0 \)
- The sum of the products of the roots taken two at a time \( \alpha\beta + \beta\gamma + \gamma\alpha = \frac{c}{a} = \frac{2}{1} = 2 \)
- The product of the roots \( \alpha\beta\gamma = -\frac{d}{a} = -\frac{-3}{1} = 3 \)

### Step 4: Calculate \( \alpha^3 + \beta^3 + \gamma^3 \)
We can use the identity:

\[ \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma \]

Since \( \alpha + \beta + \gamma = 0 \), the first term becomes zero:

\[ \alpha^3 + \beta^3 + \gamma^3 = 0 \cdot (\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma \]
\[ = 0 + 3 \cdot 3 = 9 \]

### Final Result
Thus, the value of \( \alpha^3 + \beta^3 + \gamma^3 \) is:

\[ \alpha^3 + \beta^3 + \gamma^3 = 9 \]