If `(x^(4))/((x-1)(x+2))=(1)/(3(x-1))-(16)/(3(x+2))+x^(2)-x+k` then `k=`

Options

A. 0
B. 1
C. 2
D. 3

Answer

To solve the equation

\[ \frac{x^4}{(x-1)(x+2)} = \frac{1}{3(x-1)} - \frac{16}{3(x+2)} + x^2 - x + k \]

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

### Step 1: Multiply both sides by the common denominator
Multiply both sides by \((x-1)(x+2)\) to eliminate the denominators:

\[ x^4 = \frac{1}{3}(x+2) - \frac{16}{3}(x-1) + (x^2 - x + k)(x-1)(x+2) \]

### Step 2: Simplify the right-hand side
Distributing the terms on the right-hand side:

1. For \(\frac{1}{3}(x+2)\):
\[ \frac{1}{3}(x+2) = \frac{x}{3} + \frac{2}{3} \]

2. For \(-\frac{16}{3}(x-1)\):
\[ -\frac{16}{3}(x-1) = -\frac{16x}{3} + \frac{16}{3} \]

Combining these gives:
\[ \frac{x}{3} - \frac{16x}{3} + \frac{2}{3} + \frac{16}{3} = -\frac{15x}{3} + \frac{18}{3} = -5x + 6 \]

Now we have:
\[ x^4 = -5x + 6 + (x^2 - x + k)(x-1)(x+2) \]

### Step 3: Expand \((x^2 - x + k)(x-1)(x+2)\)
First, expand \((x-1)(x+2)\):
\[ (x-1)(x+2) = x^2 + x - 2 \]

Now distribute \((x^2 - x + k)\):
\[ (x^2 - x + k)(x^2 + x - 2) = x^4 + x^3 - 2x^2 - x^3 - x^2 + 2x + kx^2 + kx - 2k \]

Combining like terms:
\[ = x^4 + (0)x^3 + (-2 - 1 + k)x^2 + (2 + k)x - 2k \]
\[ = x^4 + (k - 3)x^2 + (k + 2)x - 2k \]

### Step 4: Set the equation equal
Now we equate both sides:
\[ x^4 = -5x + 6 + x^4 + (k - 3)x^2 + (k + 2)x - 2k \]

### Step 5: Cancel \(x^4\) and simplify
Cancelling \(x^4\) from both sides:
\[ 0 = -5x + 6 + (k - 3)x^2 + (k + 2)x - 2k \]

### Step 6: Combine like terms
Combine the terms:
\[ 0 = (k - 3)x^2 + (-5 + k + 2)x + (6 - 2k) \]

### Step 7: Set coefficients equal to zero
Since this must hold for all \(x\), we set the coefficients of \(x^2\), \(x\), and the constant term to zero:

1. Coefficient of \(x^2\):
\[ k - 3 = 0 \implies k = 3 \]

2. Coefficient of \(x\):
\[ -5 + k + 2 = 0 \implies k - 3 = 0 \implies k = 3 \]

3. Constant term:
\[ 6 - 2k = 0 \implies 2k = 6 \implies k = 3 \]

### Conclusion
In all cases, we find that \( k = 3 \).

Thus, the value of \( k \) is

\[ \boxed{3} \]