Solve the following inequation .
(xiii) `(x^2+x+1)^xlt1`

Answer

To solve the inequation \((x^2 + x + 1)^x < 1\), we can follow these steps:

### Step 1: Understanding the Inequation
We need to find the values of \(x\) for which the expression \((x^2 + x + 1)^x\) is less than 1.

### Step 2: Taking Logarithm
To simplify the inequation, we can take the natural logarithm (log to the base \(e\)) on both sides:
\[ \log((x^2 + x + 1)^x) < \log(1) \]
Since \(\log(1) = 0\), we can rewrite the inequation as:
\[ x \cdot \log(x^2 + x + 1) < 0 \]

### Step 3: Analyzing the Cases
This inequation can be analyzed based on the sign of \(x\).

#### Case 1: \(x > 0\)
If \(x\) is positive, we can divide both sides of the inequation by \(x\) (since \(x > 0\), the inequality sign remains the same):
\[ \log(x^2 + x + 1) < 0 \]
This implies:
\[ x^2 + x + 1 < 1 \]
Subtracting 1 from both sides gives:
\[ x^2 + x < 0 \]
Factoring out \(x\):
\[ x(x + 1) < 0 \]
The roots of the equation \(x(x + 1) = 0\) are \(x = 0\) and \(x = -1\). The sign of the product \(x(x + 1)\) is negative in the interval \((-1, 0)\). However, since we are in the case where \(x > 0\), this case does not yield any valid solutions.

#### Case 2: \(x < 0\)
If \(x\) is negative, we again divide both sides by \(x\) (note that this reverses the inequality):
\[ \log(x^2 + x + 1) > 0 \]
This implies:
\[ x^2 + x + 1 > 1 \]
Subtracting 1 from both sides gives:
\[ x^2 + x > 0 \]
Factoring out \(x\):
\[ x(x + 1) > 0 \]
The roots of the equation \(x(x + 1) = 0\) are \(x = 0\) and \(x = -1\). The product \(x(x + 1)\) is positive in the intervals \((-\infty, -1)\) and \((0, \infty)\). However, since we are in the case where \(x < 0\), we only consider the interval \((-\infty, -1)\).

### Step 4: Conclusion
Combining the results from both cases, the solution set for the inequation \((x^2 + x + 1)^x < 1\) is:
\[ x \in (-\infty, -1) \]