Помогите решить: p^2-4q^2//(p+2q)^2 : p^3-8q^3//4q^2+2pq+p^2; 3x^2-3y^2//x^2+xy * x+y//6x-6y;

[latex] \frac{p^2-4q^2}{(p+2q)^2} = \frac{(p-2q)(p+2q)}{(p+2q)^2} =\frac{p-2q}{p+2q} [/latex] [latex] \frac{ p^3-8q^3}{4q^2+2pq+p^2} = \frac{(p-2q)(4q^2+2pq+p^2)}{4q^2+2pq+p^2} = p-2q [/latex] [latex] \frac{3x^2-3y^2}{x^2+xy} * \frac{x+y}{6x-6y} = \frac{3(x+y)^{2}(x-y) }{6x(x+y)(x-y)} = \frac{x+y}{2x} [/latex]