Помогите решить 7 и 8

Решите задачу:

$\displaystyle\tt\boxed{7}\\\\ (4x^2-2xy+y^2)-(*)=3x^2+3xy\\\\(*)=(4x^2-2xy+y^2)-(3x^2+3xy)\\\\ (*)=4x^2-2xy+y^2-3x^2-3xy\\\\ \bold{(*)=x^2-5xy+y^2}\\\\\\ \boxed{8}\\\\ 2a^2d^3=7 \ \ \Rightarrow \ \ a^2d^3=\frac{7}{2}\\\\\\ a) \ -6a^2d^3=-6\cdot\frac{7}{2} = -21\\\\\\ b) \ 2a^4d^6=2{(a^2d^3)}^2=2\cdot \bigg(\frac{7}{2}\bigg)^2=2\cdot\frac{49}{4} =\frac{49}{2} =24\frac{1}{2} =24.5$