УМОЛЯЮ решите эти два номера !!!

Решите задачу:

$\displaystyle 28. \\ a).\ \ (3a-2b)^{2}-(3a+2b)^{2}=(3a-2b-3a-2b)(3a-2b+3a+2b)= \\=-4b*6a=-24ab; \\ \\ b).\ \ 9(2x-y)^{2}-(5x-y)^{2}=(3(2x-y))^{2}-(5x-y)^{2}= \\ =(6x-3y-5x+y)(6x-3y+5x-y)=(x-2y)(11x-4y); \\ \\ c).\ \ 4(a-3b)^{2}-9(5a-4b)^{2}=(2(a-3b))^{2}-(3(5a-4b))^{2}= \\ =(2a-6b)^{2}-(15a-12b)^{2}= \\ =(2a-6b-15a+12b)(2a-6b+15a-12b)=(6b-13a)(17a-18b);$

$\displaystyle 29. \\ a).\ \ x^{2}+14x+48=x^{2}+14x+49 - 1=(x+7)^{2}-1= \\ =(x+7-1)(x+7+1)=(x+6)(x+8); \\ \\ b).\ \ a^{2}-6ab+5b^{2}=a^{2}-6ab+9b^{2}-4b^{2}=(a-3b)^{2}-(2b)^{2}= \\ =(a-3b-2b)(a-3b+2b)=(a-5b)(a-b); \\ \\ c).\ \ 4p^{4}+q^{4}=(2p^{2})^{2}+4p^{2}q^{2}+(q^{2})^{2}-4p^{2}q^{2}= \\ =(2p^{2}+q^{2})^{2}-(2pq)^{2}=(2p^{2}+q^{2}-2pq)(2p^{2}+q^{2}+2pq);$