Question

question
factor $8c^3 - 27$ completely.
answer attempt 1 out of 2

Explanation:

Step1: Identify the formula

The expression \(8c^3 - 27\) is a difference of cubes. The formula for factoring a difference of cubes is \(a^3 - b^3=(a - b)(a^2+ab + b^2)\).
Here, \(a = 2c\) because \((2c)^3=8c^3\) and \(b = 3\) because \(3^3 = 27\).

Step2: Apply the formula

Substitute \(a = 2c\) and \(b = 3\) into the difference of cubes formula:
\[

$$\begin{align*} 8c^3-27&=(2c)^3 - 3^3\\ &=(2c - 3)((2c)^2+(2c)(3)+3^2)\\ &=(2c - 3)(4c^2 + 6c+9) \end{align*}$$

\]
We check if \(4c^2 + 6c + 9\) can be factored further. The discriminant of the quadratic \(ax^2+bx + c\) is \(b^2-4ac\). For \(4c^2+6c + 9\), \(a = 4\), \(b = 6\), \(c = 9\). The discriminant is \(6^2-4\times4\times9=36 - 144=- 108<0\), so it cannot be factored over the real numbers.

Answer:

\((2c - 3)(4c^2 + 6c + 9)\)