\((a-b)^{2}=a^{2}-2 a b+b^{2}\)
put a = xyz, b = 1 inthe above identity,
\( (x y z-1)^{2} =(x y z)^{2}-2 \times x y z \times 1+1^{2} \)
\(\\ =x^{2} y^{2} z^{2}-2 x y z+1 \)

\( (a-b)^{2}=a^{2}-2 a b+b^{2}\)
\(\\ put \ a=b, b=7 \)
\(\\\text {in the above identity,} \)
\(\\ (b-7)^{2} =b^{2}-2 \times b \times 7+7^{2}\)
\( \\ =b^{2}-14 b+49 \)

(x + a)(x + b)= x² + x(a + b) + ab
put x = 6A, a = 9, b = 5 in the above formula,
\( (6 a+9)(6 a-5) =(6 a) 2+6 a(9-5)-(9 \times 5) \)
\(\\ =36 a^{2}+6 a(4)-(45) \)
\(\\ =36 a^{2}+24 a-45 \)

9x² + 30xy + 25y² = (3² x x² )+(2 x 3 x 5)x xy +(5²xy² )
= (3x)² + 2x(3x) x (5y) + (5y)²
This expression is in the form of identity a² + 2ab + b² = (a + b)².
Hence, (3x)² + 2x(3x) x (5y) + (5y)² = (3x + 5y)²
= (3x + 5y)(3x + 5y).
Therefore, the factors of 9x² + 30xy + 25y² are (3x + 5y) and (3x + 5y).

(x + 3)(x + 5)
Let us represent the expression geometrically, as shown in the Figure
In the rectangle with length (x + 5) and breadth (x + 3), we get,

Area of bigger rectangle = Area of a square + Area of three rectangles
Therefore,
(x + 3)(x + 5) = x² + 5x + 3x +15
= x² + (5 + 3)x +15
= x² + 8x + 15.
Area of bigger rectangle = Area of a square + Area of three rectangles
Therefore,
(x + 3)(x + 5) = x² + 5x + 3x + 15
= x² + (5 + 3)x + 15
 = x² + 8x + 15 .
(ii) (y + 6)(y + 8)
Let us represent the expression geometrically, as shown in Figure. In the rectangle with length (y + 6) and breadth (y + 8) units, we get, 
Area of bigger rectangle = Area of square + Area of three rectangles

(iii) 43 x 36 = (40 + 3) x (40 − 4)
We know the identity
(x + a)(x + b) = x² + x(a + b) + ab
Taking, x = 40, a = 3 and b =  –4, we get
(40 + 3)(40 − 4) = 40² + 40(3 − 4) + 3(−4)
= 1600 + 40(−1) − 1²
= 1600 − 40 − 1²
= 1600 − 52
Therefore, 43 x 36 = 1548.