Question 960963
<pre>
FOIL them all out and see:

A. {{{(8x^4+50y^3)(8x^4-50y^3)}}}{{{""=""}}}{{{64x^8-400x^4y^3+400x^4y^3-2500y^6}}}{{{""=""}}}{{{64x^8-cross(400x^4y^3)+cross(400x^4y^3)-2500y^6}}}{{{""=""}}}{{{64x^8-2500y^6}}} NO
B. {{{(8x^4+50y^4)(8x^4-50y^2)}}}{{{""=""}}}{{{64x^8-400x^4y^2+400x^4y^4-2500y^6}}} NO
C. {{{(4x^4+10y^3)(4x^4-10y^3)}}}{{{""=""}}}{{{64x^8-400x^4y^3+400x^4y^3-100y^6}}}{{{""=""}}}{{{64x^8-cross(400x^4y^3)+cross(400x^4y^3)-100y^6}}}{{{""=""}}}{{{64x^8-2500y^6}}} NO
D. {{{(4x^4+10y^4)(4x^4-10y^2)}}}{{{""=""}}}{{{16x^8-40x^4y^3+40x^4y^3-100y^6}}}{{{""=""}}}{{{16x^8-cross(40x^4y^3)+cross(40x^4y^3)-100y^6}}}{{{""=""}}}{{{16x^8-100y^6}}} YES

Even though D is the correct answer, it's not completely factored.
If fact D, although mathematically correct, is not what you should 
do first when factoring.

Here's the correct way to do the problem:

{{{16x^8-100y^6}}}

Factor out greatest common factor of 4. That's what you should do first,
always!

{{{4(4x^8-25y^6)}}}

Write {{{4x^8}}} as {{{(2x^4)^2}}} and {{{25y^6}}} as {{{(5y^3)^2}}}

{{{4((2x^4)^2-(5y^3)^2)}}} 

Factor the difference of squares.

{{{4(2x^4-5y^3)(2x^4+5y^3)}}}

That's the correct complete factorization even though 
it is not listed.

Edwin</pre>