Question

помогите помогииииттеееееееееее 40 баллов

Answer 1

$\frac{1}{u-y} -\frac{3uy}{u^3-y^3} =\frac{1}{u-y} -\frac{3uy}{(u-y)(u^2+uy+y^2)} =\\ \\ \frac{u^2+uy+y^2-3uy}{(u-y)(u^2+uy+y^2)} =\frac{u^2-2uy+y^2}{(u-y)(u^2+uy+y^2)} =\\ \\ \frac{(u-y)^2}{(u-y)(u^2+uy+y^2)} =\frac{u-y}{u^2+uy+y^2}$