三、解答题(共44分)
15. (16分)计算:
(1)$(x+3)^{2}-(x+2)(x-2)$;
(2)$(2x-y-1)(2x-y+1)$;
(3)$(x-2)(x+2)(x^{2}-4)$;
(4)$(2x+y)^{2}(2x-y)^{2}$.

15. (1)原式$= x^{2} + 6x + 9 - x^{2} + 4 = 6x + 13$. (2)原式$= [(2x - y) - 1][(2x - y) + 1] = (2x - y)^{2} - 1 = 4x^{2} - 4xy + y^{2} - 1$. (3)原式$= (x^{2} - 4)^{2} = x^{4} - 8x^{2} + 16$. (4)原式$= [(2x + y)(2x - y)]^{2} = (4x^{2} - y^{2})^{2} = 16x^{4} - 8x^{2}y^{2} + y^{4}$.

16. (8分)先化简,再求值:$(2a-b)(a-2b)-(-2a+3b)^{2}-(a+b)(a-b)$,其中$a=-2$,$b=-1$.

16. 原式$= 2a^{2} - 4ab - ab + 2b^{2} - 4a^{2} + 12ab - 9b^{2} - a^{2} + b^{2} = -3a^{2} + 7ab - 6b^{2}$. 当$a = -2$，$b = -1$时，原式$= -3×(-2)^{2} + 7×(-2)×(-1) - 6×(-1)^{2} = -4$.

解：原式$=(2a-b)(a-2b)-(-2a+3b)^{2}-(a+b)(a-b)$
$=2a^{2}-4ab-ab+2b^{2}-(4a^{2}-12ab+9b^{2})-(a^{2}-b^{2})$
$=2a^{2}-5ab+2b^{2}-4a^{2}+12ab-9b^{2}-a^{2}+b^{2}$
$=-3a^{2}+7ab-6b^{2}$.
当$a=-2$，$b=-1$时，
原式$=-3×(-2)^{2}+7×(-2)×(-1)-6×(-1)^{2}$
$=-3×4+14-6×1$
$=-12+14-6$
$=-4$.

17. (9分)已知$a^{2}+b^{2}=3$,$a+b=2$.
(1)求$ab$的值.
(2)求$(a-b)^{2}$的值.
(3)求$a^{4}+b^{4}$的值.

17. (1)$\because a + b = 2$，$\therefore (a + b)^{2} = 4$，即$a^{2} + 2ab + b^{2} = 4$. 又$\because a^{2} + b^{2} = 3$，$\therefore 3 + 2ab = 4$，$\therefore ab = \frac{1}{2}$. (2)$(a - b)^{2} = (a + b)^{2} - 4ab = 4 - 4×\frac{1}{2} = 2$. (3)$a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2a^{2}b^{2} = (a^{2} + b^{2})^{2} - 2(ab)^{2} = 3^{2} - 2×(\frac{1}{2})^{2} = \frac{17}{2}$.

18. (11分)观察下列各式:
$(a+1)(a^{2}-a+1)=a^{3}+1$;
$(a-2)(a^{2}+2a+4)=a^{3}-8$;
$(3a-2)(9a^{2}+6a+4)=27a^{3}-8$.
(1)根据上述各式的运算规律填空.
①$(x-3)(x^{2}+3x+9)=$;
②$(2x+1)$()$=8x^{3}+1$;
③()$(x^{2}+xy+y^{2})=x^{3}-y^{3}$.
(2)应用规律计算:$(a^{2}-b^{2})(a^{2}+ab+b^{2})(a^{2}-ab+b^{2})$.

18. (1)①$x^{3} - 27$ ②$4x^{2} - 2x + 1$ ③$x - y$
(2)$(a^{2} - b^{2})(a^{2} + ab + b^{2})(a^{2} - ab + b^{2}) = [(a + b)(a^{2} - ab + b^{2})][(a - b)(a^{2} + ab + b^{2})] = (a^{3} + b^{3})(a^{3} - b^{3}) = a^{6} - b^{6}$.