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 \(1 + 2r + 4r^2 \neq 0\).