Question 335893
<pre>
what is 64x^3+8y^3 divided by 8x+2y?


{{{(64x^3+8y^3)/(8x+2y)}}}

Factor 8 out of the top and 2 out of the bottom.

{{{(8(8x^3+y^3))/(2(4x+y))}}}

{{{4}}}
{{{cross(8)(8x^3+y^3)/(cross(2)(4x+y))}}}
 
{{{(4(8x^3+y^3))/(4x+y)}}}

You could factor the sum of two cubes in parentheses 
in the numerator but there is no use since nothing would 
cancel.

Are you sure the problem wasn't supposed to have been:

"What is 64x^3+8y^3 divided by <font color = "red">4</font>x+2y?"
instead of "by <font color = "red">8</font>x+2y", for then you could 
factor the sum of cubes and and cancel. 

Edwin</pre>