Find the domain of definitions of the following function: `f(x)=(x^(2)+x+1)^(-3//2)`

Explanation

To find the domain of the function \( f(x) = (x^2 + x + 1)^{-3/2} \), we need to determine the values of \( x \) for which the function is defined.

### Step-by-Step Solution:

1. **Rewrite the Function**:
The function can be rewritten as:
\[ f(x) = \frac{1}{(x^2 + x + 1)^{3/2}} \]
This shows that the function involves a denominator, which means we need to ensure that the denominator is not zero.

2. **Set Conditions for the Denominator**:
For \( f(x) \) to be defined, the expression in the denominator must be positive:
\[ (x^2 + x + 1)^{3/2} \neq 0 \]
This implies:
\[ x^2 + x + 1 \neq 0 \]

3. **Determine When the Expression is Non-Negative**:
Additionally, since we are dealing with a square root, we need:
\[ x^2 + x + 1 > 0 \]

4. **Analyze the Quadratic Expression**:
The quadratic expression \( x^2 + x + 1 \) can be analyzed using the discriminant:
\[ \text{Discriminant} = b^2 - 4ac = 1^2 - 4 \cdot 1 \cdot 1 = 1 - 4 = -3 \]
Since the discriminant is negative, the quadratic has no real roots and does not cross the x-axis.

5. **Determine the Sign of the Quadratic**:
The coefficient of \( x^2 \) is positive (1), which means the parabola opens upwards. Therefore, \( x^2 + x + 1 \) is always positive for all real numbers \( x \).

6. **Conclusion on the Domain**:
Since \( x^2 + x + 1 > 0 \) for all \( x \in \mathbb{R} \), the function \( f(x) \) is defined for all real numbers.

Thus, the domain of the function \( f(x) \) is:
\[ \text{Domain of } f(x) = \mathbb{R} \]