Question

If \( \frac{x^3+3}{(x-3)^3} = a + \frac{b}{x-3} + \frac{c}{(x-3)^2} + \frac{d}{(x-3)^3} \), then \( (a+d)-(b+c) = \)

• A. 49
• B. 15
• C. -30
• D. -5

Hint:

When the denominator is of the form \( (x-k)^n \), substitution \( t=x-k \) is faster than using derivatives or long division to find coefficients.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:

The problem involves decomposing a rational function into partial fractions. Since the denominator contains a repeated linear factor \( (x-3)^3 \), we can use a substitution method to express the numerator as a polynomial in terms of \( (x-3) \).
Step 2: Key Formula or Approach:

Substitute \( t = x - 3 \), which implies \( x = t + 3 \). This transforms the expression into a polynomial divided by a monomial, simplifying the decomposition.
Step 3: Detailed Explanation:

Let \( x - 3 = t \). Then \( x = t + 3 \).
Substitute \( x \) into the numerator \( x^3 + 3 \): \[ (t+3)^3 + 3 = (t^3 + 3t^2(3) + 3t(3)^2 + 27) + 3 \] \[ = t^3 + 9t^2 + 27t + 27 + 3 \] \[ = t^3 + 9t^2 + 27t + 30 \] Now, divide by the denominator \( (x-3)^3 = t^3 \): \[ \frac{t^3 + 9t^2 + 27t + 30}{t^3} = \frac{t^3}{t^3} + \frac{9t^2}{t^3} + \frac{27t}{t^3} + \frac{30}{t^3} \] \[ = 1 + \frac{9}{t} + \frac{27}{t^2} + \frac{30}{t^3} \] Substitute \( t = x-3 \) back into the expression: \[ 1 + \frac{9}{x-3} + \frac{27}{(x-3)^2} + \frac{30}{(x-3)^3} \] Comparing this with the given form: \[ a = 1, \quad b = 9, \quad c = 27, \quad d = 30 \] We need to calculate \( (a+d) - (b+c) \): \[ (1 + 30) - (9 + 27) = 31 - 36 = -5 \]
Step 4: Final Answer:

The value is -5.