If `a_(1), a_(2), a_(3).... A_(n) in R^(+) and a_(1).a_(2).a_(3).... A_(n) = 1`, then minimum value of `(1 + a_(1) + a_(1)^(2)) (a + a_(2) + a_(2)^(2)) (1 + a_(3) + a_(3)^(2))..... (1 + a_(n) + a_(n)^(2))` is equal to

Options

A. `3^(n + 1)`
B. `3^(n)`
C. `3^(n - 1)`
D. none of these

Answer

To solve the problem, we need to find the minimum value of the expression:

\[ (1 + a_1 + a_1^2)(1 + a_2 + a_2^2)(1 + a_3 + a_3^2) \ldots (1 + a_n + a_n^2) \]

given that \(a_1, a_2, a_3, \ldots, a_n \in \mathbb{R}^+\) and \(a_1 \cdot a_2 \cdot a_3 \cdots a_n = 1\).

### Step 1: Analyze the expression

We notice that each term in the product can be expressed as:

\[ 1 + a_i + a_i^2 \]

### Step 2: Use AM-GM Inequality

To minimize \(1 + a_i + a_i^2\), we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. According to AM-GM:

\[ \frac{1 + a_i + a_i^2}{3} \geq \sqrt[3]{1 \cdot a_i \cdot a_i^2} \]

This simplifies to:

\[ \frac{1 + a_i + a_i^2}{3} \geq \sqrt[3]{a_i^3} = a_i \]

Thus, we can rearrange this to find:

\[ 1 + a_i + a_i^2 \geq 3a_i \]

### Step 3: Substitute for \(a_i\)

Since we have the constraint \(a_1 \cdot a_2 \cdots a_n = 1\), we can set \(a_i = x\) for all \(i\) (i.e., \(a_1 = a_2 = \ldots = a_n = x\)). Therefore, we have:

\[ x^n = 1 \implies x = 1 \]

### Step 4: Calculate the minimum value

Substituting \(x = 1\) back into the expression:

\[ 1 + 1 + 1^2 = 1 + 1 + 1 = 3 \]

Thus, the product becomes:

\[ (1 + 1 + 1^2)^n = 3^n \]

### Conclusion

The minimum value of the expression is:

\[ \boxed{3^n} \]