Question

In the following exercises, factor completely.

$8x^3-27y^3$

Short answer

The expression is factored as$8x^3-27y^3=(2x-3y)(4x^2+6xy+9y^2)$

Step-by-step solution

Step 1. Given Information

Consider the expression$8x^3-27y^3$

The objective is to factor the expression completely.

Step 2. Factor using the difference of cube.

The difference of cube can be factored as

$a^3-b^3=(a-b)(a^2+ab+b^2)$

In the given binomial $8x^3-27y^3$, the first term $8x^3$is the cube of $2x$and the second term $27y^3$is the cube of $3y$. So it can be factored using the difference of cube formula as

localid="1644651551111" $\begin{array}{rcl}8x^3-27y^3& =& {\left(2x\right)}^3-{\left(3y\right)}^3\\ & =& (2x-3y)\{{\left(2x\right)}^2+(2x\left)\right(3y)+{\left(3y\right)}^2\}\\ & =& (2x-3y)(4x^2+6xy+9y^2)\end{array}$

Step 3. Check the factors

Multiply the factors to check the solution

$\begin{array}{rcl}(2x-3y)(4x^2+6xy+9y^2)& =& 2x(4x^2+6xy+9y^2)-3y(4x^2+6xy+9y^2)\\ & =& 8x^3+12x^{2}y+18x{y}^2-12x^{2}y-18x{y}^2-27y^3\\ & =& 8x^3-27y^3\end{array}$

And we get the given expression. So the expression is correctly factored.