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To determine whether the expression [tex]\(\frac{x^3 - 1}{x^2 - 1}\)[/tex] simplifies to [tex]\(x\)[/tex], let's conduct a step-by-step simplification of the expression.
### Step 1: Factor the Numerator and Denominator
First, we need to factor both the numerator ([tex]\(x^3 - 1\)[/tex]) and the denominator ([tex]\(x^2 - 1\)[/tex]).
Factor the numerator:
The expression [tex]\(x^3 - 1\)[/tex] can be factored using the difference of cubes formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
In our case, [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \][/tex]
Factor the denominator:
The expression [tex]\(x^2 - 1\)[/tex] can be factored using the difference of squares formula:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
In our case, [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
### Step 2: Rewrite the Expression with Factored Forms
Substitute the factored forms of the numerator and denominator into the original expression:
[tex]\[ \frac{x^3 - 1}{x^2 - 1} = \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} \][/tex]
### Step 3: Simplify the Expression
We can now cancel the common factor [tex]\((x - 1)\)[/tex] in the numerator and the denominator (assuming [tex]\(x \neq 1\)[/tex]):
[tex]\[ \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{x^2 + x + 1}{x + 1} \][/tex]
### Step 4: Examine the Simplified Expression
The simplified expression is [tex]\(\frac{x^2 + x + 1}{x + 1}\)[/tex]. It does not simplify further to [tex]\(x\)[/tex].
To determine whether [tex]\(\frac{x^2 + x + 1}{x + 1} = x\)[/tex], you can examine it more closely:
For the expression [tex]\(\frac{x^2 + x + 1}{x + 1}\)[/tex] to equal [tex]\(x\)[/tex]:
[tex]\[ x \neq -1 \][/tex]
both sides of the equation [tex]\(\frac{x^2 + x + 1}{x + 1} = x\)[/tex] should hold:
Set up the equation:
[tex]\[ \frac{x^2 + x + 1}{x + 1} = x \][/tex]
Cross multiply to solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 + x + 1 = x \cdot (x + 1) \][/tex]
[tex]\[ x^2 + x + 1 = x^2 + x \][/tex]
Subtract [tex]\(x^2 + x\)[/tex] from both sides:
[tex]\[ 1 = 0 \][/tex]
### Conclusion
Clearly, the above equation [tex]\(1 = 0\)[/tex] is a contradiction, which means [tex]\(\frac{x^2 + x + 1}{x + 1} \neq x\)[/tex].
So, [tex]\(\frac{x^3 - 1}{x^2 - 1} \neq x\)[/tex].
Therefore, the correct answer is:
No, because the simplified form of the expression [tex]\(\frac{x^3 - 1}{x^2 - 1}\)[/tex] is [tex]\(\frac{x^2 + x + 1}{x + 1}\)[/tex] and it doesn't simplify to [tex]\(x\)[/tex].
### Step 1: Factor the Numerator and Denominator
First, we need to factor both the numerator ([tex]\(x^3 - 1\)[/tex]) and the denominator ([tex]\(x^2 - 1\)[/tex]).
Factor the numerator:
The expression [tex]\(x^3 - 1\)[/tex] can be factored using the difference of cubes formula:
[tex]\[ a^3 - b^3 = (a - b)(a^2 + ab + b^2) \][/tex]
In our case, [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^3 - 1 = (x - 1)(x^2 + x + 1) \][/tex]
Factor the denominator:
The expression [tex]\(x^2 - 1\)[/tex] can be factored using the difference of squares formula:
[tex]\[ a^2 - b^2 = (a - b)(a + b) \][/tex]
In our case, [tex]\(a = x\)[/tex] and [tex]\(b = 1\)[/tex]:
[tex]\[ x^2 - 1 = (x - 1)(x + 1) \][/tex]
### Step 2: Rewrite the Expression with Factored Forms
Substitute the factored forms of the numerator and denominator into the original expression:
[tex]\[ \frac{x^3 - 1}{x^2 - 1} = \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} \][/tex]
### Step 3: Simplify the Expression
We can now cancel the common factor [tex]\((x - 1)\)[/tex] in the numerator and the denominator (assuming [tex]\(x \neq 1\)[/tex]):
[tex]\[ \frac{(x - 1)(x^2 + x + 1)}{(x - 1)(x + 1)} = \frac{x^2 + x + 1}{x + 1} \][/tex]
### Step 4: Examine the Simplified Expression
The simplified expression is [tex]\(\frac{x^2 + x + 1}{x + 1}\)[/tex]. It does not simplify further to [tex]\(x\)[/tex].
To determine whether [tex]\(\frac{x^2 + x + 1}{x + 1} = x\)[/tex], you can examine it more closely:
For the expression [tex]\(\frac{x^2 + x + 1}{x + 1}\)[/tex] to equal [tex]\(x\)[/tex]:
[tex]\[ x \neq -1 \][/tex]
both sides of the equation [tex]\(\frac{x^2 + x + 1}{x + 1} = x\)[/tex] should hold:
Set up the equation:
[tex]\[ \frac{x^2 + x + 1}{x + 1} = x \][/tex]
Cross multiply to solve for [tex]\(x\)[/tex]:
[tex]\[ x^2 + x + 1 = x \cdot (x + 1) \][/tex]
[tex]\[ x^2 + x + 1 = x^2 + x \][/tex]
Subtract [tex]\(x^2 + x\)[/tex] from both sides:
[tex]\[ 1 = 0 \][/tex]
### Conclusion
Clearly, the above equation [tex]\(1 = 0\)[/tex] is a contradiction, which means [tex]\(\frac{x^2 + x + 1}{x + 1} \neq x\)[/tex].
So, [tex]\(\frac{x^3 - 1}{x^2 - 1} \neq x\)[/tex].
Therefore, the correct answer is:
No, because the simplified form of the expression [tex]\(\frac{x^3 - 1}{x^2 - 1}\)[/tex] is [tex]\(\frac{x^2 + x + 1}{x + 1}\)[/tex] and it doesn't simplify to [tex]\(x\)[/tex].
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