Let `alpha,beta,gamma` are the real roots of the equation `x^3+a x^2+b x+c=0(a ,b ,c in Ra n da!=0)dot` If the system of equations `(inu ,v ,a n d w)` given by `alphau+betav+gammaw=0` `betau+gammav+alphaw=0` `gammau+alphav+betaw=0` has non-trivial solutions then the value of `a^2//b` is ________.

Explanation

To solve the problem, we need to analyze the given cubic equation and the system of equations. Let's break it down step by step.

### Step 1: Understand the cubic equation
The cubic equation given is:
\[ x^3 + ax^2 + bx + c = 0 \]
where \( \alpha, \beta, \gamma \) are the roots of this equation.

### Step 2: Use Vieta's formulas
From Vieta's formulas, we know:
1. \( \alpha + \beta + \gamma = -a \)
2. \( \alpha\beta + \beta\gamma + \gamma\alpha = b \)
3. \( \alpha\beta\gamma = -c \)

### Step 3: Write down the system of equations
The system of equations is:
1. \( \alpha u + \beta v + \gamma w = 0 \)
2. \( \beta u + \gamma v + \alpha w = 0 \)
3. \( \gamma u + \alpha v + \beta w = 0 \)

### Step 4: Set up the determinant
For the system to have non-trivial solutions, the determinant of the coefficients must be zero. We can write the determinant as:
\[ \begin{vmatrix} \alpha & \beta & \gamma \\ \beta & \gamma & \alpha \\ \gamma & \alpha & \beta \end{vmatrix} = 0 \]

### Step 5: Expand the determinant
We can expand this determinant along the first row:
\[ D = \alpha \begin{vmatrix} \gamma & \alpha \\ \alpha & \beta \end{vmatrix} - \beta \begin{vmatrix} \beta & \alpha \\ \gamma & \beta \end{vmatrix} + \gamma \begin{vmatrix} \beta & \gamma \\ \gamma & \alpha \end{vmatrix} \]

Calculating the 2x2 determinants, we have:
1. \( \begin{vmatrix} \gamma & \alpha \\ \alpha & \beta \end{vmatrix} = \gamma \beta - \alpha^2 \)
2. \( \begin{vmatrix} \beta & \alpha \\ \gamma & \beta \end{vmatrix} = \beta^2 - \alpha\gamma \)
3. \( \begin{vmatrix} \beta & \gamma \\ \gamma & \alpha \end{vmatrix} = \beta \alpha - \gamma^2 \)

Substituting these back into the determinant:
\[ D = \alpha(\gamma \beta - \alpha^2) - \beta(\beta^2 - \alpha\gamma) + \gamma(\beta \alpha - \gamma^2) \]

### Step 6: Simplify the determinant
After simplifying, we arrive at:
\[ D = \alpha \gamma \beta - \alpha^3 - \beta^3 + \alpha \beta \gamma - \gamma^3 = 0 \]
This can be rearranged to:
\[ \alpha^3 + \beta^3 + \gamma^3 - 3\alpha\beta\gamma = 0 \]

### Step 7: Use the identity for cubes
Using the identity for the sum of cubes:
\[ \alpha^3 + \beta^3 + \gamma^3 = (\alpha + \beta + \gamma)(\alpha^2 + \beta^2 + \gamma^2 - \alpha\beta - \beta\gamma - \gamma\alpha) \]
Substituting \( \alpha + \beta + \gamma = -a \) and \( \alpha\beta + \beta\gamma + \gamma\alpha = b \):
\[ \alpha^3 + \beta^3 + \gamma^3 = -a \left( \frac{a^2}{3} - b \right) \]

### Step 8: Set the equation to zero
Setting the equation to zero gives us:
\[ -a \left( \frac{a^2}{3} - b \right) = 3\alpha\beta\gamma \]
Substituting \( \alpha\beta\gamma = -c \) leads to:
\[ -a \left( \frac{a^2}{3} - b \right) = -3c \]

### Step 9: Solve for \( \frac{a^2}{b} \)
From the equation \( a^2 = 3b \), we can write:
\[ \frac{a^2}{b} = 3 \]

### Final Answer
Thus, the value of \( \frac{a^2}{b} \) is:
\[ \boxed{3} \]