Question

StartFraction 1 minus 8 r cubed Over 12 r squared plus 6 r plus 3 EndFraction

Answer 1

To simplify the expression

\[
\frac{1 - 8r^3}{12r^2 + 6r + 3},
\]

we begin by examining both the numerator and the denominator.

### Step 1: Factor the Numerator
The numerator \(1 - 8r^3\) can be factored using the difference of cubes formula:

\[
a^3 - b^3 = (a - b)(a^2 + ab + b^2).
\]

Here, we can rewrite \(1\) as \(1^3\) and \(8r^3\) as \((2r)^3\). Thus:

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

### Step 2: Factor the Denominator
Now, let’s factor the denominator \(12r^2 + 6r + 3\). First, we can factor out the common factor of \(3\):

\[
12r^2 + 6r + 3 = 3(4r^2 + 2r + 1).
\]

Next, we will check if the quadratic \(4r^2 + 2r + 1\) can be factored further. We can see that its discriminant \(b^2 - 4ac = 2^2 - 4 \cdot 4 \cdot 1 = 4 - 16 = -12\), which is negative, indicating that the quadratic does not factor nicely over the real numbers.

### Step 3: Combine the Results
Now we combine the factored forms:

\[
\frac{1 - 8r^3}{12r^2 + 6r + 3} = \frac{(1 - 2r)(1 + 2r + 4r^2)}{3(4r^2 + 2r + 1)}.
\]

Since \(4r^2 + 2r + 1\) cannot be simplified further and does not factor, we can’t cancel any terms between the numerator and denominator. Thus the simplified form remains:

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

This is the simplified expression.