If `alpha,beta,gamma` are the roots of the equation `x^3+a_0x^2+a_1x+a_2=0`, then `(1-alpha^2)(1-beta^2)(1-gamma^2)` is eqyal to:

Options

A. `(1-a_1)^2+(a_0-a_2)^2`
B. `(1+a_1)^2-(a_0-a_2)^2`
C. `(1+a_1)^2+(a_0+a_2)^2`
D. none of these

Answer

To find the value of \((1 - \alpha^2)(1 - \beta^2)(1 - \gamma^2)\) given that \(\alpha, \beta, \gamma\) are the roots of the cubic equation \(x^3 + a_0 x^2 + a_1 x + a_2 = 0\), we can follow these steps:

### Step 1: Expand the Expression
We start with the expression:
\[ (1 - \alpha^2)(1 - \beta^2)(1 - \gamma^2) \]
We can expand this using the distributive property:
\[ = 1 - (\alpha^2 + \beta^2 + \gamma^2) + (\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2) - \alpha^2\beta^2\gamma^2 \]

### Step 2: Use Vieta's Formulas
From Vieta's formulas for the roots of the polynomial \(x^3 + a_0 x^2 + a_1 x + a_2 = 0\), we know:
- \(\alpha + \beta + \gamma = -a_0\)
- \(\alpha\beta + \beta\gamma + \gamma\alpha = a_1\)
- \(\alpha\beta\gamma = -a_2\)

### Step 3: Calculate \(\alpha^2 + \beta^2 + \gamma^2\)
We can express \(\alpha^2 + \beta^2 + \gamma^2\) in terms of the sums of the roots:
\[ \alpha^2 + \beta^2 + \gamma^2 = (\alpha + \beta + \gamma)^2 - 2(\alpha\beta + \beta\gamma + \gamma\alpha) \]
Substituting the values from Vieta's:
\[ = (-a_0)^2 - 2a_1 = a_0^2 - 2a_1 \]

### Step 4: Calculate \(\alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2\)
This can be expressed as:
\[ \alpha^2\beta^2 + \beta^2\gamma^2 + \gamma^2\alpha^2 = (\alpha\beta + \beta\gamma + \gamma\alpha)^2 - 2\alpha\beta\gamma(\alpha + \beta + \gamma) \]
Substituting the values:
\[ = a_1^2 - 2(-a_2)(-a_0) = a_1^2 - 2a_2a_0 \]

### Step 5: Calculate \(\alpha^2\beta^2\gamma^2\)
This is simply:
\[ \alpha^2\beta^2\gamma^2 = (\alpha\beta\gamma)^2 = (-a_2)^2 = a_2^2 \]

### Step 6: Substitute Everything Back
Now we substitute back into our expanded expression:
\[ (1 - \alpha^2)(1 - \beta^2)(1 - \gamma^2) = 1 - (a_0^2 - 2a_1) + (a_1^2 - 2a_2a_0) - a_2^2 \]
Simplifying this gives:
\[ = 1 - a_0^2 + 2a_1 + a_1^2 - 2a_2a_0 - a_2^2 \]

### Final Expression
Thus, the final expression for \((1 - \alpha^2)(1 - \beta^2)(1 - \gamma^2)\) is:
\[ 1 - a_0^2 + 2a_1 + a_1^2 - 2a_2a_0 - a_2^2 \]