Fill in the blanks.

Solution:

(i) $(x-9)^{2}$

$x^{2}-18 x+81$

$=x^{2}-2 \times x \times 9+9^{2}$

$=(x-9)^{2}$

(ii) (4) $(1-3 x)(1+3 x)$

$4-36 x^{2}$

$=4\left(1-9 x^{2}\right)$

$=4\left(1^{2}-(3 x)^{2}\right)$

$=(4)(1-3 x)(1+3 x)$

(iii) $(x-13)(x-1)$

$x^{2}-14 x+13$

$=x^{2}-13 x-x+13$

$=x(x-13)-1(x-13)$

$=(x-13)(x-1)$

(iv) $(3 z+x-2 y)(3 z-x+2 y)$

$9 z^{2}-x^{2}-4 y^{2}+4 x y$

$=9 z^{2}-\left(x^{2}-4 x y+4 y^{2}\right)$

$=9 z^{2}-\left(x^{2}-2 \times x \times 2 y+(2 y)^{2}\right)$

$=9 z^{2}-(x-2 y)^{2}$

$=(3 z)^{2}-(x-2 y)^{2}$

$=(3 z+(x-2 y))(3 z-(x-2 y))$

$=(3 z+x-2 y)(3 z-x+2 y)$

(v) $(c-1)(a b-1)$

$a b c-a b-c+1$

$=a b(c-1)-1(c-1)$

$=(c-1)(a b-1)$