To address why the x^3 cancels out in Example 1 but not in Example 2, we need to examine the operations involved in each example.
Example 1:
\[ \frac{(x^3 - 8)}{(x - 2)} = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)} \]
In this example, we can simplify by canceling out the common factor of (x - 2), leaving us with \( x^2 + 2x + 4 \).
Example 2:
\[ \frac{(x^3 - 8)}{(x - 2)} = \frac{(x - 2)(x^2 + 2x + 4)}{(x - 2)} \]
Here, it seems like the intention is to cancel out the (x - 2) factors, but the cancellation isn't correctly executed. We can't simply cancel out the entire (x - 2) term from both the numerator and denominator. This is an incorrect application of canceling out terms.
The difference lies in how the cancellation is performed. In Example 1, we cancel out a common factor (x - 2) from both the numerator and the denominator, which is valid. However, in Example 2, attempting to cancel out the entire (x - 2) term from both numerator and denominator is incorrect because it's not a factor that can be directly canceled out like a common term.
