Question

factoring diff / sum of cubes (level 1)
this is the only question in this section.
question
factor ( x^3 + 27q^3 ) completely.
answer attempt 1 out of 2

Explanation:

Step1: Identify sum of cubes

$p^3 + 27q^3 = p^3 + (3q)^3$

Step2: Apply sum of cubes formula

Recall $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$. Let $a=p$, $b=3q$.

$$\begin{align*} p^3 + (3q)^3 &= (p+3q)(p^2 - p(3q) + (3q)^2)\\ &=(p+3q)(p^2 - 3pq + 9q^2) \end{align*}$$

Step3: Verify no further factoring

The quadratic $p^2 - 3pq + 9q^2$ cannot be factored over integers.

Answer:

$(p + 3q)(p^2 - 3pq + 9q^2)$