Notice that you have a difference of cubes:
[tex]\dfrac{x^3}8-\dfrac{y^3}{27}=\left(\dfrac x2\right)^3-\left(\dfrac y3\right)^3[/tex]
which can be factored using the rule,
[tex]a^3-b^3=(a-b)(a^2+ab+b^2)[/tex]
So,
[tex]\dfrac{x^3}8-\dfrac{y^3}{27}=\left(\dfrac x2-\dfrac y3\right)\left(\dfrac{x^2}4+\dfrac{xy}6+\dfrac{y^2}9\right)[/tex]
and just to make things look nicer, we can pull out 1/6 from the first factor and 1/36 from the second to get
[tex]\dfrac{x^3}8-\dfrac{y^3}{27}=\dfrac1{216}(3x-2y)(9x^2+6xy+4y^2)[/tex]
