Answer :
Let's analyze and simplify the given function step by step.
Given:
[tex]\[ f(x) = \frac{x^3 - 1}{x^3 + 1} \][/tex]
### Step 1: Factorize the numerator and the denominator.
First, note that both the numerator [tex]\(x^3 - 1\)[/tex] and the denominator [tex]\(x^3 + 1\)[/tex] can be factored using standard algebraic identities.
The difference of cubes identity is:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
The sum of cubes identity is:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Using these identities, we can factorize [tex]\(x^3 - 1\)[/tex] and [tex]\(x^3 + 1\)[/tex]:
For [tex]\(x^3 - 1\)[/tex], let [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x \cdot 1 + 1^2) = (x - 1)(x^2 + x + 1) \][/tex]
For [tex]\(x^3 + 1\)[/tex], let [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^3 + 1 = (x + 1)(x^2 - x \cdot 1 + 1^2) = (x + 1)(x^2 - x + 1) \][/tex]
### Step 2: Substitute the factored forms back into the function.
Now, substituting these factored forms into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{(x - 1)(x^2 + x + 1)}{(x + 1)(x^2 - x + 1)} \][/tex]
### Step 3: Simplify the expression.
Upon inspecting the numerator and the denominator, we observe that [tex]\(x^2 + x + 1\)[/tex] and [tex]\(x^2 - x + 1\)[/tex] do not simplify further through cancellation. The function, therefore, simplifies to:
[tex]\[ f(x) = \frac{(x - 1)(x^2 + x + 1)}{(x + 1)(x^2 - x + 1)} \][/tex]
Thus, the simplified function is already suitable for verifying behaviors such as limits, asymptotes, or further analysis.
### Final Answer
The function [tex]\( f(x) \)[/tex] is represented as:
[tex]\[ f(x) = \frac{x^3 - 1}{x^3 + 1} \][/tex]
This expression accurately describes the given function [tex]\( f(x) \)[/tex] in its factored form, which confirms our step-by-step simplification.
Given:
[tex]\[ f(x) = \frac{x^3 - 1}{x^3 + 1} \][/tex]
### Step 1: Factorize the numerator and the denominator.
First, note that both the numerator [tex]\(x^3 - 1\)[/tex] and the denominator [tex]\(x^3 + 1\)[/tex] can be factored using standard algebraic identities.
The difference of cubes identity is:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
The sum of cubes identity is:
[tex]\[ a^3 + b^3 = (a + b)(a^2 - ab + b^2) \][/tex]
Using these identities, we can factorize [tex]\(x^3 - 1\)[/tex] and [tex]\(x^3 + 1\)[/tex]:
For [tex]\(x^3 - 1\)[/tex], let [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x \cdot 1 + 1^2) = (x - 1)(x^2 + x + 1) \][/tex]
For [tex]\(x^3 + 1\)[/tex], let [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^3 + 1 = (x + 1)(x^2 - x \cdot 1 + 1^2) = (x + 1)(x^2 - x + 1) \][/tex]
### Step 2: Substitute the factored forms back into the function.
Now, substituting these factored forms into the function [tex]\( f(x) \)[/tex]:
[tex]\[ f(x) = \frac{(x - 1)(x^2 + x + 1)}{(x + 1)(x^2 - x + 1)} \][/tex]
### Step 3: Simplify the expression.
Upon inspecting the numerator and the denominator, we observe that [tex]\(x^2 + x + 1\)[/tex] and [tex]\(x^2 - x + 1\)[/tex] do not simplify further through cancellation. The function, therefore, simplifies to:
[tex]\[ f(x) = \frac{(x - 1)(x^2 + x + 1)}{(x + 1)(x^2 - x + 1)} \][/tex]
Thus, the simplified function is already suitable for verifying behaviors such as limits, asymptotes, or further analysis.
### Final Answer
The function [tex]\( f(x) \)[/tex] is represented as:
[tex]\[ f(x) = \frac{x^3 - 1}{x^3 + 1} \][/tex]
This expression accurately describes the given function [tex]\( f(x) \)[/tex] in its factored form, which confirms our step-by-step simplification.
