Question

factoring diff / sum of cubes (level 1)
score: 1/2 penalty: none
question
factor $y^3 - 1$ completely.

Explanation:

Step1: Recall the formula for difference of cubes

The formula for factoring the difference of cubes is $a^3 - b^3=(a - b)(a^2+ab + b^2)$.

Step2: Identify \(a\) and \(b\) in the given expression

In the expression \(y^3 - 1\), we can rewrite \(1\) as \(1^3\). So, \(a = y\) and \(b = 1\) since \(y^3-1=y^3 - 1^3\).

Step3: Apply the difference of cubes formula

Substitute \(a = y\) and \(b = 1\) into the formula \(a^3 - b^3=(a - b)(a^2+ab + b^2)\).
We get \((y - 1)(y^2+(y\times1)+1^2)=(y - 1)(y^2 + y + 1)\).

Answer:

$(y - 1)(y^2 + y + 1)$