If `(x^(4))/((x-1)(x-2)(x-3))= A.x+B. (1)/((x-1))+C (1)/((x-2))+D. (1)/((x-3))+E`, then `A+B+C+D+E=`

Options

A. -12
B. 6
C. 18
D. 32

Answer

To solve the equation

\[ \frac{x^4}{(x-1)(x-2)(x-3)} = A \cdot x + \frac{B}{(x-1)} + \frac{C}{(x-2)} + \frac{D}{(x-3)} + E, \]

we need to express the left-hand side as a sum of partial fractions on the right-hand side. Let's break this down step by step.

### Step 1: Rewrite the equation
We start by rewriting the right-hand side with a common denominator:

\[ A \cdot x + \frac{B}{(x-1)} + \frac{C}{(x-2)} + \frac{D}{(x-3)} + E = \frac{A \cdot x \cdot (x-1)(x-2)(x-3) + B \cdot (x-2)(x-3) + C \cdot (x-1)(x-3) + D \cdot (x-1)(x-2) + E \cdot (x-1)(x-2)(x-3)}{(x-1)(x-2)(x-3)}. \]

### Step 2: Eliminate the denominator
By multiplying both sides by \((x-1)(x-2)(x-3)\), we get:

\[ x^4 = A \cdot x \cdot (x-1)(x-2)(x-3) + B \cdot (x-2)(x-3) + C \cdot (x-1)(x-3) + D \cdot (x-1)(x-2) + E \cdot (x-1)(x-2)(x-3). \]

### Step 3: Substitute values for x
To find the coefficients \(A\), \(B\), \(C\), \(D\), and \(E\), we can substitute specific values for \(x\).

1. **Substituting \(x = 1\)**:
\[ 1^4 = A \cdot 1 \cdot 0 + B \cdot (1-2)(1-3) + 0 + 0 + 0. \]
This simplifies to:
\[ 1 = B \cdot (-1)(-2) \implies 1 = 2B \implies B = \frac{1}{2}. \]

2. **Substituting \(x = 2\)**:
\[ 2^4 = A \cdot 2 \cdot 0 + 0 + C \cdot (2-1)(2-3) + 0 + 0. \]
This simplifies to:
\[ 16 = C \cdot 1 \cdot (-1) \implies 16 = -C \implies C = -16. \]

3. **Substituting \(x = 3\)**:
\[ 3^4 = A \cdot 3 \cdot 0 + 0 + 0 + D \cdot (3-1)(3-2) + 0. \]
This simplifies to:
\[ 81 = D \cdot 2 \cdot 1 \implies 81 = 2D \implies D = \frac{81}{2}. \]

### Step 4: Find \(A\) and \(E\)
To find \(A\) and \(E\), we can equate coefficients. The left-hand side is \(x^4\), and we need to find the coefficient of \(x^3\) on the right-hand side.

The coefficient of \(x^3\) from the right-hand side is given by:

\[ A + B + C + D + E. \]

Since \(x^4\) has no \(x^3\) term, we can set the sum of coefficients equal to 0:

\[ A + \frac{1}{2} - 16 + \frac{81}{2} + E = 0. \]

### Step 5: Solve for \(A\) and \(E\)
We can simplify this equation:

\[ A + \frac{1}{2} - 16 + \frac{81}{2} + E = 0 \implies A + E + \frac{1 + 81 - 32}{2} = 0 \implies A + E + \frac{50}{2} = 0 \implies A + E + 25 = 0. \]

Thus,

\[ A + E = -25. \]

### Step 6: Find \(A + B + C + D + E\)
Now we can find \(A + B + C + D + E\):

\[ A + \frac{1}{2} - 16 + \frac{81}{2} + E = 0. \]

Substituting \(A + E = -25\):

\[ -25 + \frac{1}{2} - 16 + \frac{81}{2} = -25 + 40 = 15. \]

Thus,

\[ A + B + C + D + E = 15. \]

### Final Answer
The final result is:

\[ A + B + C + D + E = 15. \]