## anonymous 4 years ago multiply the sum of 4a/a-2b and 8b/a+2b by a^2-4b^2/4

$(\frac{4a}{a-2b}+\frac{8b}{a+2b})(\frac{a^2-4b^2}{4})=(\frac{4a(a+2b)+8b(a-2b)}{(a-2b)(a+2b)})(\frac{(a-2b)(a+2b)}{4})$ $(\frac{4a^2+16ab-16b^2}{(a-2b)(a+2b)})(\frac{(a-2b)(a+2b)}{4})=\frac{4(a^2+4ab-4b^2)}{4}=a^2+4ab-4b^2$ Difference of 2 squares: a^2-4b^2=(a-2b)(a+2b)