Question

question 1-10
factor completely: ( y^3 - 27 )
( (y - 3)^3 )
( (y - 3)(y^2 + 3y + 9) )
( (y - 3)(y + 3)^2 )
( (y - 6)(y^2 + 12y + 9) )

Explanation:

Step1: Recall the difference of cubes formula

The formula for factoring the difference of two 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 \(y^3-27\), we can rewrite \(27\) as \(3^3\). So, \(a = y\) and \(b = 3\) since \(y^3-27=y^3 - 3^3\).

Step3: Apply the difference of cubes formula

Substitute \(a = y\) and \(b = 3\) into the formula \((a - b)(a^2+ab + b^2)\). We get \((y - 3)(y^2+(y)(3)+3^2)=(y - 3)(y^2 + 3y+9)\).

Answer:

\((y - 3)(y^2 + 3y + 9)\) (corresponding to the option \((y - 3)(y^2+3y + 9)\))