Factor completely.
$y^3-\frac{1}{27}$

• A. $(y+\frac{1}{3})(y^2+\frac{1}{3}y+\frac{1}{9})$
• B. $(y-\frac{1}{3})(y^2+\frac{1}{3}y+\frac{1}{9})$
• C. $y{(y-\frac{1}{3})}^2$
• D. $-y{(y+\frac{1}{3})}^2$

Answer: (B)

Verified step by step guidance

1. Identify whether the given expression is a sum or difference of cubes. The general forms are \(a^3 + b^3\) for sum of cubes and \(a^3 - b^3\) for difference of cubes.
2. Rewrite the expression in the form \(a^3 \pm b^3\) by recognizing the cube roots of each term. For example, if the term is \$8x^3\(, rewrite it as \)(2x)^3\( because \)2^3 = 8$.
3. Use the appropriate factoring formula: for sum of cubes, use \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\); for difference of cubes, use \(a^3 - b^3 = (a - b)(a^2 + ab + b^2)\).
4. Substitute the values of \(a\) and \(b\) (the cube roots found in step 2) into the formula and write out the factored form explicitly.
5. Simplify each term inside the parentheses if possible, and check your work by expanding the factors to ensure they multiply back to the original expression.