Question

factoring diff / sum of cubes (level 1)
score: 0/2 penalty: none
question
factor $27b^3 + 1$ completely.
answer

Explanation:

Step1: Identify the sum of cubes

Recall the sum of cubes formula: $a^3 + b^3=(a + b)(a^2 - ab + b^2)$. For $27b^3+1$, we can rewrite $27b^3$ as $(3b)^3$ and $1$ as $1^3$. So, $a = 3b$ and $b = 1$ in the sum of cubes formula.

Step2: Apply the sum of cubes formula

Substitute $a = 3b$ and $b = 1$ into the formula $a^3 + b^3=(a + b)(a^2 - ab + b^2)$.
First, $a + b=3b + 1$.
Then, $a^2=(3b)^2 = 9b^2$, $ab=(3b)(1)=3b$, and $b^2 = 1^2 = 1$. So, $a^2 - ab + b^2=9b^2 - 3b + 1$.
Thus, $27b^3 + 1=(3b + 1)(9b^2 - 3b + 1)$.

Answer:

$(3b + 1)(9b^2 - 3b + 1)$