Question

(1-8r^3) / (1+2r+4r^2)

Answer 1

To simplify the expression \((1-8r^3) / (1+2r+4r^2)\), we can first check if either the numerator or the denominator can be factored.

### Step 1: Factoring the Numerator
The numerator \(1 - 8r^3\) can be recognized as a difference of cubes:

\[
1 - 8r^3 = 1 - (2r)^3
\]

Using the difference of cubes factorization \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\), where \(a = 1\) and \(b = 2r\):

\[
1 - 8r^3 = (1 - 2r)(1^2 + 1 \cdot 2r + (2r)^2)
\]
\[
= (1 - 2r)(1 + 2r + 4r^2)
\]

### Step 2: Factoring the Denominator
The denominator \(1 + 2r + 4r^2\) is already in the form that corresponds to the quadratic formula, or it could be checked for simple factoring:

\[
1 + 2r + 4r^2
\]

### Step 3: Combining the Factors
Now we can substitute the factorization of the numerator into the expression:

\[
\frac{1 - 8r^3}{1 + 2r + 4r^2} = \frac{(1 - 2r)(1 + 2r + 4r^2)}{1 + 2r + 4r^2}
\]

### Step 4: Cancelling Common Factors
Since \(1 + 2r + 4r^2\) is a common factor in both the numerator and denominator, we can cancel this out (assuming \(1 + 2r + 4r^2 \neq 0\)):

\[
= 1 - 2r
\]

### Final Result
Thus, the simplified form of \(\frac{1 - 8r^3}{1 + 2r + 4r^2}\) is:

\[
\boxed{1 - 2r}
\]

Keep in mind that this simplification is valid as long as \(1 + 2r + 4r^2 \neq 0\).