Question 74321
Notice how we are dealing with perfect squares here. So the equation factors to 
{{{(6y+5)(6y-5)}}} which foils back to 
{{{(6y+5)(6y-5)=6y*6y+cross(6y*5-6y*5)-5*5=36y^2-25}}}
In a more general sense
{{{x^2-y^2}}} factors to {{{(x+y)(x-y)}}} or in other words
{{{x^2-y^2=(x+y)(x-y)}}} which is the difference of squares where x=6 and y=5
So the answer is a). 
Hope this will help you with further difference of squares problems.