Question 111367
For this problem we need to use the difference of squares and the difference of cubes formulas. 


Remember the difference of squares formula is:

{{{x^2-y^2=(x+y)(x-y)}}}


and the difference of cubes formula is:

{{{x^3-y^3=(x-y)(x^2+xy+y^2)}}}



So rewrite the numerator {{{a^3-64}}} as {{{(a)^3-(4)^3}}} notice how {{{x=a}}} and {{{y=b}}} for the formula {{{x^2-y^2=(x+y)(x-y)}}}. Now factor {{{(a)^3-(4)^3}}} to {{{(a-4)(a^2+4a+16)}}}


So our expression now looks like this: {{{(a-4)(a^2+4a+16)/(a^2-16)}}}


Now factor the denominator using the difference of squares


{{{(a-4)(a^2+4a+16)/((a+4)(a-4))}}}



Cancel like terms


{{{cross((a-4))(a^2+4a+16)/((a+4)cross((a-4)))}}}



Simplify


{{{(a^2+4a+16)/(a+4)}}}


So this shows that {{{(a^3-64)/(a^2-16)}}} simplifies to {{{(a^2+4a+16)/(a+4)}}}