Question

question 1 of 5
select the correct answer.
what is the factored form of $8x^3 - 27y^3$?
$\bigcirc$ $(2x + 3y)(4x^2 - 6xy + 9y^2)$
$\bigcirc$ $(2x - 3y)(4x^2 + 9xy + 6y^2)$
$\bigcirc$ $(2x - 3y)(4x^2 + 6xy + 9y^2)$
$\bigcirc$ $(2x + 3y)(4x^2 - 9xy + 6y^2)$

Explanation:

Step1: Recall the difference of cubes formula

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

Step2: Identify \(a\) and \(b\) in \(8x^3 - 27y^3\)

We can rewrite \(8x^3\) as \((2x)^3\) and \(27y^3\) as \((3y)^3\). So, \(a = 2x\) and \(b=3y\).

Step3: Apply the difference of cubes formula

Substitute \(a = 2x\) and \(b = 3y\) into the formula \((a - b)(a^2+ab + b^2)\).
First, \(a - b=2x - 3y\).
Then, \(a^2=(2x)^2 = 4x^2\), \(ab=(2x)(3y)=6xy\), and \(b^2=(3y)^2 = 9y^2\). So, \(a^2+ab + b^2=4x^2+6xy + 9y^2\).
Putting it together, \(8x^3 - 27y^3=(2x - 3y)(4x^2+6xy + 9y^2)\).

Answer:

\((2x - 3y)(4x^2 + 6xy + 9y^2)\) (the third option: \((2x - 3y)(4x^2 + 6xy + 9y^2)\))