Solve ` (1/2)^(x^(6)-2x^(4)) lt 2^((x)^(2))`.

Explanation

To solve the inequality \( \left( \frac{1}{2} \right)^{x^6 - 2x^4} < 2^{x^2} \), we can follow these steps:

### Step 1: Rewrite the base
We start by rewriting the left-hand side using the property of exponents:
\[ \left( \frac{1}{2} \right)^{x^6 - 2x^4} = 2^{-(x^6 - 2x^4)} = 2^{-(x^6 - 2x^4)}. \]
Thus, the inequality becomes:
\[ 2^{-(x^6 - 2x^4)} < 2^{x^2}. \]

### Step 2: Compare the exponents
Since the bases are the same (both are base 2), we can compare the exponents:
\[ -(x^6 - 2x^4) < x^2. \]
This simplifies to:
\[ - x^6 + 2x^4 < x^2. \]

### Step 3: Rearrange the inequality
Rearranging gives:
\[ -x^6 + 2x^4 - x^2 < 0. \]
Multiplying through by -1 (which reverses the inequality) gives:
\[ x^6 - 2x^4 + x^2 > 0. \]

### Step 4: Factor the expression
We can factor out \( x^2 \):
\[ x^2 (x^4 - 2x^2 + 1) > 0. \]
The quadratic \( x^4 - 2x^2 + 1 \) can be rewritten as:
\[ x^4 - 2x^2 + 1 = (x^2 - 1)^2. \]
Thus, we have:
\[ x^2 (x^2 - 1)^2 > 0. \]

### Step 5: Analyze the factors
The expression \( x^2 (x^2 - 1)^2 > 0 \) is positive when:
1. \( x^2 > 0 \) (which is true for all \( x \neq 0 \)).
2. \( (x^2 - 1)^2 > 0 \) (which is true for all \( x \neq 1 \) and \( x \neq -1 \)).

### Step 6: Combine the conditions
Thus, the solution to the inequality is:
\[ x \in (-\infty, -1) \cup (-1, 0) \cup (0, 1) \cup (1, \infty). \]

### Final Answer
The solution set is:
\[ x \in (-\infty, -1) \cup (-1, 0) \cup (0, 1) \cup (1, \infty). \]