91Ó°ÊÓ

The expression in the denominator, \( x^3 + 27 \), is a classic example of a sum of cubes. A sum of cubes is any expression that takes the form \( a^3 + b^3 \). This type of expression can be factored using the identity \( a^3 + b^3 = (a+b)(a^2 - ab + b^2) \).

- In our exercise, \( a = x \) and \( b = 3 \), since \( 3^3 = 27 \).
- Applying the formula, we rewrite \( x^3 + 27 \) as \((x + 3)(x^2 - 3x + 9) \).

Recognizing and factoring such expressions is essential in simplifying them for further operations, such as partial fraction decomposition.

Factorization is the process of breaking down an expression into a product of simpler expressions, or "factors," which when multiplied together give the original expression. In this case, recognizing the sum of cubes allows us to factor \( x^3 + 27 \) into \((x + 3)(x^2 - 3x + 9) \).

This step is crucial as it transforms a complex denominator into factors that can be more easily managed in a partial fraction decomposition. It allows us to separate the given complex rational expression into simpler fractions, making it easier to work with. Factorization remains a key strategy in mathematical problem-solving, especially in calculus and algebra.

In order to find the partial fraction decomposition, we need to find the constants \( A, B, \) and \( C \). This requires setting up and solving a system of equations.

To begin, we multiply through by the common denominator \((x + 3)(x^2 - 3x + 9)\) of the rational equation, leading to:
- \( -5x^2 + 6x - 2 = A(x^2 - 3x + 9) + (Bx + C)(x + 3) \)

This expansion then gives us equations in terms of the coefficients of like powers of \( x \). Each of these results in a separate linear equation. Solving this system involves substituting and manipulating equations to find values for \( A, B, \) and \( C \).

The method of equating coefficients is used to solve for the constants in a partial fraction decomposition once the expressions are expanded. This involves:

- Matching the coefficients of corresponding powers of \( x \) on both sides of the equation.
- For this exercise, the coefficients are equated as follows:
- For \(x^2\): \( A + B = -5 \)
- For \(x\): \( -3A + 3B + C = 6 \)
- For the constant term: \( 9A + 3C = -2 \)

This method simplifies finding solutions for multiple variables within polynomial identities. Solving these equations helps determine the values of \( A, B, \) and \( C \), which are then used in the final decomposition. This systematic approach is critical in algebra and calculus, often providing a way to break down and understand complex equations.