Answered

Does the expression $\frac{x^3-1}{x^2-1}$ simplify to $x$?


ANo, because $x^3-1$ can be factored as $x\left(x^2-x+1\right)$ and $x^2-1$ can be factored as $x(x-1)$, so only $x$ can be canceled.
BYes, because $x^3(x-1)$ can be factored as $x^2(x-1)$ and $x^2-1$ can be factored as $x(x-1)$, so $(x-1)$ can be canceled.
CNo, because the -1 in the numerator and denominator is not a common factor and cannot be canceled.
DYes, because -1 in the numerator and denominator is a common factor and can be canceled.

Verified Answer :

  • Factor the numerator \(x^3-1\) into \((x-1)(x^2+x+1)\) and the denominator \(x^2-1\) into \((x-1)(x+1)\).
  • Simplify the fraction to \(\frac{x^2+x+1}{x+1}\) for \(x eq 1\).
  • Compare the simplified expression with \(x\), observing it is actually \(x + \frac{1}{x+1}\).
  • Conclude the answer is: No, because the \(-1\) in the numerator and denominator is not a common factor and cannot be canceled. \(\boxed{\text{No, because the -1 in the numerator and denominator is not a common factor and cannot be canceled.}}\)

Explanation

  1. Factoring the Expression

The expression given is \(\frac{x^3-1}{x^2-1}\). We want to determine if this simplifies to \(x\). To analyze this, we start by factoring both the numerator and the denominator using standard algebraic identities. The numerator \(x^3 - 1\) is a difference of cubes, which follows the identity \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\). Setting \(a = x\) and \(b = 1\), we get:
\[x^3 - 1 = (x - 1)(x^2 + x + 1).\]
The denominator \(x^2 - 1\) is a difference of squares, which follows the identity \(a^2 - b^2 = (a - b)(a + b)\). Setting \(a = x\) and \(b = 1\), we get:
\[x^2 - 1 = (x - 1)(x + 1).\]

  1. Simplifying and Comparing to x

Now we substitute these factored forms back into the original expression:
\[\frac{x^3-1}{x^2-1} = \frac{(x-1)(x^2+x+1)}{(x-1)(x+1)}.\]
For all \(x eq 1\), we can cancel the common factor of \((x - 1)\) from the numerator and denominator, which leaves us with:
\[\frac{x^2+x+1}{x+1}.\]
To determine if this is equivalent to \(x\), we compare it to \(x = \frac{x(x+1)}{x+1} = \frac{x^2+x}{x+1}\). Since \(\frac{x^2+x+1}{x+1} = \frac{x^2+x}{x+1} + \frac{1}{x+1} = x + \frac{1}{x+1}\), the expression is clearly not equal to \(x\).

  1. Identifying the Correct Reasoning

A common algebraic mistake is attempting to "cancel" parts of an expression that are not factors of the entire numerator or denominator. For example, some might try to cancel the \(-1\) in the numerator with the \(-1\) in the denominator. However, \(-1\) is a term being subtracted, not a factor, and thus it cannot be canceled. The correct conclusion is that the expression does not simplify to \(x\). Among the choices provided, the one that correctly identifies the error in attempting to cancel terms is the third option.

Examples

When working with algebraic fractions, such as in electrical circuit analysis or signal processing, incorrectly canceling terms can lead to significant errors in calculations. For instance, if you are simplifying a transfer function like \(\frac{s^2+s+1}{s+1}\), knowing that you cannot simply cancel terms allows you to maintain the integrity of your model, ensuring that the resulting behavior of the system matches the physical reality of the circuit.

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