At Westonci.ca, we connect you with the answers you need, thanks to our active and informed community. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Explore comprehensive solutions to your questions from a wide range of professionals on our user-friendly platform.
Sagot :
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].
Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. We appreciate your time. Please revisit us for more reliable answers to any questions you may have. We're glad you visited Westonci.ca. Return anytime for updated answers from our knowledgeable team.
