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

Options

A. `(x+1)/(x^(2)-x+1)+(x-1)/(x^(2)+x-1)`
B. `(x-1)/(x^(2)-x+1)-(x+1)/(x^(2)+x-1)`
C. `(x)/(x^(2)-x+1)+(x+1)/(x^(2)+x-1)`
D. `(1)/(x^(2)-x+1)-(1)/(x^(2)+x+1)`

Answer

To solve the equation \(\frac{2x}{x^4 + x^2 + 1}\) using partial fractions, we will follow these steps:

### Step 1: Factor the Denominator
First, we need to factor the denominator \(x^4 + x^2 + 1\). We can rewrite it as:
\[ x^4 + x^2 + 1 = (x^2)^2 + (x^2) + 1 \]
Let \(y = x^2\), then it becomes \(y^2 + y + 1\). This quadratic does not factor nicely over the reals, but we can use the roots of unity or complete the square to find its factors.

The roots of \(y^2 + y + 1 = 0\) are:
\[ y = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Thus, we can express \(x^4 + x^2 + 1\) as:
\[ (x^2 - \frac{-1 + i\sqrt{3}}{2})(x^2 - \frac{-1 - i\sqrt{3}}{2}) \]
However, for our purposes, we can express it as:
\[ (x^2 - x + 1)(x^2 + x + 1) \]

### Step 2: Set Up Partial Fractions
Now, we can express \(\frac{2x}{x^4 + x^2 + 1}\) in terms of partial fractions:
\[ \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{Ax + B}{x^2 - x + 1} + \frac{Cx + D}{x^2 + x + 1} \]

### Step 3: Combine the Right Side
To combine the right side, we take the common denominator:
\[ \frac{(Ax + B)(x^2 + x + 1) + (Cx + D)(x^2 - x + 1)}{(x^2 - x + 1)(x^2 + x + 1)} \]
This gives us:
\[ Ax^3 + Ax^2 + Ax + Bx^2 + Bx + B + Cx^3 - Cx^2 + Cx + Dx^2 - Dx + D \]
Combining like terms, we have:
\[ (A + C)x^3 + (A + B - C + D)x^2 + (A + B + C - D)x + (B + D) \]

### Step 4: Set Up Equations
Now, we can set up equations by comparing coefficients from both sides:
1. For \(x^3\): \(A + C = 0\)
2. For \(x^2\): \(A + B - C + D = 0\)
3. For \(x\): \(A + B + C - D = 2\)
4. For the constant term: \(B + D = 0\)

### Step 5: Solve the System of Equations
From equation 1, we have \(C = -A\). Substituting into the other equations:
- From equation 4: \(D = -B\)
- Substitute \(C\) and \(D\) into equation 2:
\[ A + B - (-A) - B = 0 \implies 2A = 0 \implies A = 0 \implies C = 0 \]
- Substitute \(A\) and \(C\) into equation 3:
\[ 0 + B + 0 - (-B) = 2 \implies 2B = 2 \implies B = 1 \implies D = -1 \]

### Step 6: Write the Partial Fraction Decomposition
Now we can write the partial fractions:
\[ \frac{2x}{(x^2 - x + 1)(x^2 + x + 1)} = \frac{0 \cdot x + 1}{x^2 - x + 1} + \frac{0 \cdot x - 1}{x^2 + x + 1} \]
This simplifies to:
\[ \frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1} \]

### Final Result
Thus, the partial fraction decomposition of \(\frac{2x}{x^4 + x^2 + 1}\) is:
\[ \frac{1}{x^2 - x + 1} - \frac{1}{x^2 + x + 1} \]