f.3 maths　簡易多項式的因式分解 急]]

1
(a) 81 - u^4
= (9 + u^2)(9 - u^2)
= (9 + u^2)(3 + u)(3 - u)
By comparing both sides,
A = 9, B = 1, C = 3, D = 1

(b) P = 3 - u

2.
4 - 20(c - 3d) + 25(c - 3d)^2
= (2)^2 - 2 (2) [5(c - 3d)] + [5(c - 3d)]^2
= [(2) - 5(c - 3d)]^2
= (2 - 5c + 15d)^2

3.
(a - b)^2 + 6(a - b) + 9
= (a - b)^2 + 2 (a - b) (3) + (3)^2
= [(a - b) + (3)]^2
= (a - b + 3)^2

4.
(a) 1 - m^3n^3
= 1^3 - (mn)^3
= (1 - mn)[1 + mn + (mn)^2]
By comparing both sides,
A = 1
B = -1

(b) P = 1 + mn + m^2n^2

5.
(a) 185y^3 + 320(x^3 - y^3)
= 320x^3 - 135y^3
= 5 (64x^3 - 27y^3)
= 5 [(4x)^3 - (3y)^3]
= 5 (4x - 3y)[(4x)^2 + (4x)(3y) + (3y)^2]
= 5 (4x - 3y) (16x^2 + 12xy + 9y^2)
By comparing both sides,
A = 5
B = 4
C = -3

(b) P = (16x^2 + 12xy + 9y^2)

The answer of 1is:
(a) 81 - u^4
= (9 + u^2)(9 - u^2)
= (9 + u^2)(3 + u)(3 - u)
By comparing both sides,
A = 9, B = 1, C = 3, D = 1

(b) P = 3 - u

The answer of 2 is:
4 - 20(c - 3d) + 25(c - 3d)^2
= (2)^2 - 2 (2) [5(c - 3d)] + [5(c - 3d)]^2
= [(2) - 5(c - 3d)]^2
= (2 - 5c + 15d)^2

The answer of 3 is:
(a - b)^2 + 6(a - b) + 9
= (a - b)^2 + 2 (a - b) (3) + (3)^2
= [(a - b) + (3)]^2
= (a - b + 3)^2

The answer of 4 is:
(a) 1 - m^3n^3
= 1^3 - (mn)^3
= (1 - mn)[1 + mn + (mn)^2]
By comparing both sides,
A = 1
B = -1

(b) P = 1 + mn + m^2n^2

The answer of 5 is:
(a) 185y^3 + 320(x^3 - y^3)
= 320x^3 - 135y^3
= 5 (64x^3 - 27y^3)
= 5 [(4x)^3 - (3y)^3]
= 5 (4x - 3y)[(4x)^2 + (4x)(3y) + (3y)^2]
= 5 (4x - 3y) (16x^2 + 12xy + 9y^2)
By comparing both sides,
A = 5
B = 4
C = -3

(b) P = (16x^2 + 12xy + 9y^2)