Problem 87 - Introductory Algebra for College Students

Question

Perform the indicated operation or operations. Simplify the result, if possible. $$\frac{x+6}{x^{3}-27}-\frac{x}{x^{3}+3 x^{2}+9 x}$$

Step-by-Step Solution

Answer

The simplified form of the expression \(\frac{x+6}{x^{3}-27}-\frac{x}{x^{3}+3 x^{2}+9 x}\) is \(\frac{9}{(x^{2}+3x+9)(x-3)}\).

Step 1: Simplify the denominators

Firstly, simplify the denominator of each fraction. The denominator of the first fraction \(x^{3}-27\) can be factored as \((x-3)(x^{2}+3x+9)\), an application of the difference of cubes. The denominator of the second fraction is already in its simplest form \(x^{3}+3 x^{2}+9 x\). However, we can factor out an \(x\) to match the factored form of the first denominator, which means the denominator should be expressed as \(x(x^{2}+3x+9)\).

Step 2: Find the common denominator and subtract the fractions

To subtract the two fractions, they need to have the same denominator. The common denominator here is \(x(x-3)(x^{2}+3x+9)\). Therefore, the first fraction should be multiplied by \(x/x\) and the second fraction should be multiplied by \((x-3)/(x-3)\). This results in: \[\frac{x(x+6)}{x(x-3)(x^{2}+3x+9)} - \frac{(x-3)x}{x(x-3)(x^{2}+3x+9)}\] Next, subtract the numerators and write the result over the common denominator. This gives: \[\frac{x^{2}+6x - x^{2} +3x}{x(x^{2}+3x+9)(x-3)} = \frac{9x}{x(x^{2}+3x+9)(x-3)}\]

Step 3: Simplify the result

The numerator and the denominator can be further simplified by factoring out \(x\) from both the numerator and the denominator. This results in: \[\frac{9}{(x^{2}+3x+9)(x-3)}\]