If `(1-x+6x^(2))/(x-x^(3))=A/x +(B)/(1-x)+(C)/(1+x)`, then `A=`

Options

A. 1
B. 2
C. 3
D. 4

Answer

To solve the equation

\[ \frac{1 - x + 6x^2}{x - x^3} = \frac{A}{x} + \frac{B}{1 - x} + \frac{C}{1 + x} \]

we need to find the value of \( A \).

### Step 1: Factor the denominator on the left side
The denominator \( x - x^3 \) can be factored as:

\[ x(1 - x)(1 + x) \]

So we rewrite the left side:

\[ \frac{1 - x + 6x^2}{x(1 - x)(1 + x)} \]

### Step 2: Set up the equation
Now, we can equate the two sides:

\[ \frac{1 - x + 6x^2}{x(1 - x)(1 + x)} = \frac{A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x)}{x(1 - x)(1 + x)} \]

### Step 3: Clear the denominators
Since the denominators are the same, we can focus on the numerators:

\[ 1 - x + 6x^2 = A(1 - x)(1 + x) + Bx(1 + x) + Cx(1 - x) \]

### Step 4: Expand the right side
Now, we expand the right side:

1. \( A(1 - x)(1 + x) = A(1 - x^2) = A - Ax^2 \)
2. \( Bx(1 + x) = Bx + Bx^2 \)
3. \( Cx(1 - x) = Cx - Cx^2 \)

Combining these gives:

\[ A - Ax^2 + Bx + Bx^2 + Cx - Cx^2 = A + (B + C)x + (-A + B - C)x^2 \]

### Step 5: Combine like terms
Now we can combine like terms:

\[ 1 - x + 6x^2 = A + (B + C)x + (-A + B - C)x^2 \]

### Step 6: Equate coefficients
Now we equate coefficients from both sides:

1. Constant term: \( A = 1 \)
2. Coefficient of \( x \): \( B + C = -1 \)
3. Coefficient of \( x^2 \): \( -A + B - C = 6 \)

### Step 7: Solve for \( A \)
From the first equation, we already have:

\[ A = 1 \]

### Conclusion
Thus, the value of \( A \) is:

\[ \boxed{1} \]