Simplify (x^3-729)/(x-9)
The problem you are presenting is an algebraic exercise that requires simplifying a rational expression. Specifically, the expression given is a fraction with a polynomial numerator, \( x^3 - 729 \), and a polynomial denominator, \( x - 9 \). The task is to reduce this expression to its simplest form, potentially by factoring the numerator, identifying common factors in the numerator and denominator, and then canceling out these common factors to simplify the expression as much as possible.
$\frac{x^{3} - 729}{x - 9}$
Answer
Solution:
Step 1: Factor the numerator
Step 1.1: Express $729$ as a cube of $9$, that is $9^3$. Thus, the expression becomes $\frac{x^3 - 9^3}{x - 9}$.
Step 1.2: Recognize that $x^3$ and $9^3$ are cubes and apply the difference of cubes formula: $a^3 - b^3 = (a - b)(a^2 + ab + b^2)$, where $a = x$ and $b = 9$. This gives us $\frac{(x - 9)(x^2 + 9x + 81)}{x - 9}$.
Step 1.3: Simplify the expression.
Step 1.3.1: Rearrange the terms to place $9$ before $x$ in the binomial, resulting in $\frac{(x - 9)(x^2 + 9x + 81)}{x - 9}$.
Step 1.3.2: Compute $9^2$ to get $81$. The expression remains $\frac{(x - 9)(x^2 + 9x + 81)}{x - 9}$.
Step 2: Eliminate the common factor
Step 2.1: Remove the common factor $(x - 9)$ from the numerator and denominator, simplifying to $\frac{\cancel{(x - 9)}(x^2 + 9x + 81)}{\cancel{x - 9}}$.
Step 2.2: The remaining expression is $x^2 + 9x + 81$.
Knowledge Notes:
To solve the given problem, we employ several mathematical concepts and techniques:
Difference of Cubes: This is a special factoring formula used for expressions of the form $a^3 - b^3$, which factors into $(a - b)(a^2 + ab + b^2)$.
Simplifying Fractions: When a common factor is present in both the numerator and the denominator of a fraction, it can be canceled out to simplify the fraction.
Perfect Cubes: Recognizing that a number is a perfect cube (such as $729 = 9^3$) is essential for applying the difference of cubes formula.
Algebraic Manipulation: Rearranging terms and simplifying expressions are common techniques in algebra used to solve equations and simplify expressions.
By applying these concepts, we can simplify the given algebraic fraction to its simplest form.
