## Solving the Equation (x^2 + 1)^2 + 3x(x^2 + 1) + 2x^2 = 0

This equation appears complex, but we can solve it using a clever substitution and factorization. Here's how:

### 1. Substitution

Let's simplify the equation by substituting a new variable. We'll let **y = x^2 + 1**. Now our equation becomes:

**y^2 + 3xy + 2x^2 = 0**

### 2. Factoring

The equation now resembles a quadratic equation in terms of 'y'. We can factor it:

**(y + x)(y + 2x) = 0**

### 3. Back-Substituting

Now we substitute back the original value of 'y':

**(x^2 + 1 + x)(x^2 + 1 + 2x) = 0**

### 4. Solving for x

For the product of two terms to be zero, at least one of the terms must be zero. This gives us two separate equations:

**x^2 + x + 1 = 0****x^2 + 2x + 1 = 0**

We can solve these quadratic equations using the quadratic formula:

**x = [-b ± √(b^2 - 4ac)] / 2a**

For the first equation:

- a = 1, b = 1, c = 1
**x = [-1 ± √(1 - 4)] / 2 = (-1 ± √-3) / 2**- This results in complex solutions:
**x = (-1 ± i√3) / 2**

For the second equation:

- a = 1, b = 2, c = 1
**x = [-2 ± √(4 - 4)] / 2 = (-2 ± √0) / 2 = -1**

### 5. Solution

Therefore, the solutions to the original equation are:

**x = (-1 + i√3) / 2****x = (-1 - i√3) / 2****x = -1**