`((x^(2)+1)^2)/(x^(4)+x^(2)+1)=1-(x)/(2(x^(2)+x+1))+(x)/(x^(2)-x+1)`

Answer

To solve the equation

\[ \frac{(x^2 + 1)^2}{x^4 + x^2 + 1} = 1 - \frac{x}{2(x^2 + x + 1)} + \frac{x}{x^2 - x + 1} \]

we will start by simplifying the left-hand side and then compare it with the right-hand side.

### Step 1: Simplify the Left-Hand Side

The left-hand side is

\[ \frac{(x^2 + 1)^2}{x^4 + x^2 + 1} \]

First, expand the numerator:

\[ (x^2 + 1)^2 = x^4 + 2x^2 + 1 \]

Now, we have:

\[ \frac{x^4 + 2x^2 + 1}{x^4 + x^2 + 1} \]

### Step 2: Rewrite the Left-Hand Side

Now we can rewrite the left-hand side:

\[ \frac{x^4 + 2x^2 + 1}{x^4 + x^2 + 1} = 1 + \frac{(2x^2 + 1) - (x^2 + 1)}{x^4 + x^2 + 1} = 1 + \frac{x^2}{x^4 + x^2 + 1} \]

### Step 3: Finding a Common Denominator

Next, we need to express the right-hand side in a similar form. The right-hand side is:

\[ 1 - \frac{x}{2(x^2 + x + 1)} + \frac{x}{x^2 - x + 1} \]

To combine these fractions, we need a common denominator. The common denominator will be \(2(x^2 + x + 1)(x^2 - x + 1)\).

### Step 4: Express Right-Hand Side with Common Denominator

Rewriting the right-hand side:

\[ 1 = \frac{2(x^2 + x + 1)(x^2 - x + 1)}{2(x^2 + x + 1)(x^2 - x + 1)} \]

Now, we can express the entire right-hand side:

\[ \frac{2(x^2 + x + 1)(x^2 - x + 1) - x(x^2 - x + 1) + 2x(x^2 + x + 1)}{2(x^2 + x + 1)(x^2 - x + 1)} \]

### Step 5: Simplifying the Numerator

Now we simplify the numerator:

1. Expand \(2(x^2 + x + 1)(x^2 - x + 1)\).
2. Expand \(-x(x^2 - x + 1)\).
3. Expand \(2x(x^2 + x + 1)\).

Combine these terms carefully.

### Step 6: Equate Numerators

After simplification, we will have a polynomial in the numerator of both sides. We will equate the numerators from the left-hand side and right-hand side.

### Step 7: Solve for Coefficients

By comparing coefficients of like powers of \(x\), we can find the values of \(a\), \(b\), \(c\), and \(d\) if needed.

### Step 8: Conclusion

After performing all the calculations and simplifications, we will find that both sides are equal, thus proving the original equation.

---