If `alpha,beta and gamma` are roots of equation `x^(3)-3x^(2)+1=0`, then the value of `((alpha)/(1+alpha))^(3)+((beta)/(1+beta))^(3)+((gamma)/(1+gamma))^(3)` is __________.

Explanation

To solve the problem, we need to find the value of

\[ \left(\frac{\alpha}{1+\alpha}\right)^3 + \left(\frac{\beta}{1+\beta}\right)^3 + \left(\frac{\gamma}{1+\gamma}\right)^3 \]

where \(\alpha\), \(\beta\), and \(\gamma\) are the roots of the polynomial equation

\[ x^3 - 3x^2 + 1 = 0. \]

### Step 1: Identify the roots of the polynomial

The roots of the polynomial \(x^3 - 3x^2 + 1 = 0\) are denoted as \(\alpha\), \(\beta\), and \(\gamma\). We can use Vieta's formulas to find the sums and products of the roots:

- \(\alpha + \beta + \gamma = 3\)
- \(\alpha\beta + \beta\gamma + \gamma\alpha = 0\)
- \(\alpha\beta\gamma = -1\)

### Step 2: Define new variables for the roots

Let:

\[ \alpha_1 = \frac{\alpha}{1+\alpha}, \quad \beta_1 = \frac{\beta}{1+\beta}, \quad \gamma_1 = \frac{\gamma}{1+\gamma} \]

We need to find \(\alpha_1^3 + \beta_1^3 + \gamma_1^3\).

### Step 3: Find the polynomial whose roots are \(\alpha_1\), \(\beta_1\), and \(\gamma_1\)

To find the polynomial whose roots are \(\alpha_1\), \(\beta_1\), and \(\gamma_1\), we can perform a substitution in the original polynomial. We replace \(x\) with \(\frac{y}{1-y}\) (where \(y = \alpha_1\)).

### Step 4: Substitute and simplify

Substituting \(x = \frac{y}{1-y}\) into the polynomial:

\[ \left(\frac{y}{1-y}\right)^3 - 3\left(\frac{y}{1-y}\right)^2 + 1 = 0 \]

This leads to:

\[ \frac{y^3}{(1-y)^3} - 3\frac{y^2}{(1-y)^2} + 1 = 0 \]

Multiplying through by \((1-y)^3\) to eliminate the denominator gives:

\[ y^3 - 3y^2(1-y) + (1-y)^3 = 0 \]

### Step 5: Expand and combine like terms

Expanding the equation:

\[ y^3 - 3y^2 + 3y^3 - 1 + 3y^2 - 3y = 0 \]

This simplifies to:

\[ 3y^3 - 3y + 1 = 0 \]

### Step 6: Find the sums of roots

From the polynomial \(3y^3 - 3y + 1 = 0\), we can use Vieta's formulas again:

- The sum of the roots \(\alpha_1 + \beta_1 + \gamma_1 = 0\)
- The sum of the products of the roots taken two at a time \(\alpha_1\beta_1 + \beta_1\gamma_1 + \gamma_1\alpha_1 = -1\)
- The product of the roots \(\alpha_1\beta_1\gamma_1 = -\frac{1}{3}\)

### Step 7: Use the identity for cubes

Using the identity:

\[ \alpha_1^3 + \beta_1^3 + \gamma_1^3 - 3\alpha_1\beta_1\gamma_1 = (\alpha_1 + \beta_1 + \gamma_1)\left(\alpha_1^2 + \beta_1^2 + \gamma_1^2 - \alpha_1\beta_1 - \beta_1\gamma_1 - \gamma_1\alpha_1\right) \]

Since \(\alpha_1 + \beta_1 + \gamma_1 = 0\), we have:

\[ \alpha_1^3 + \beta_1^3 + \gamma_1^3 = 3\alpha_1\beta_1\gamma_1 \]

Substituting the value of \(\alpha_1\beta_1\gamma_1 = -\frac{1}{3}\):

\[ \alpha_1^3 + \beta_1^3 + \gamma_1^3 = 3 \left(-\frac{1}{3}\right) = -1 \]

### Final Answer

Thus, the value of

\[ \left(\frac{\alpha}{1+\alpha}\right)^3 + \left(\frac{\beta}{1+\beta}\right)^3 + \left(\frac{\gamma}{1+\gamma}\right)^3 = -1. \]