Factoring algebra II..

anyone know how to do these types of problems..

factor [sum & difference of two cubes]
x^3 + 64
x^3 - 100
27x^3 + 1
8x^3 - y^3

factor [difference of two squares]
d^2 - 36
1 - 81x^2
16a^2 - 49c^2
x^10 - 1
16x^2y^2 - 81z^2

perfect square trinomials :
4x^2 + 12x +9
4x^2 + 28x + 49
( i don't fully understand how to factor these when there is a number infront of x^2)

factoring using x^3 instead of x^2

125a^3 - 8
x^3 - 25x
x^3 - 125
x^3 + 10x^2 + 24
x^5 + 18x^4 + 81x^3

you don't have to help with all problems
but any you DO know how to do would sure help me.
& i would love to return the favor by answering your questions.. just leave a link
thanks.

1).

Note: cubert(x) = x^(1/3) = x to the cube root

Rule: x^3-y^3 = (x-y)(x^2+xy+y^2) x^3+y^3 = (x+y)(x^2-xy+y^2)

a). x^3+64 = x^3+4^3 = (x-4)(x^2+4x+16)
b). x^3-100 = (x - cubert(100))(x^2 + cubert(100)(x) + (cubert(100))^2)
c). 27x^3+1 = (3x)^3+1^3 = (3x+1)(9x^2-3x+1)
d). 8x^3-y^3 = (cubert(8)(x))^3-y^3) = (xcubert(8)-y)((xcubert(8))^2+xycubert(8…

2).

Rule: a^2-b^2 = (a-b)(a+b)

a). d^2-36 = d^2-6^2 = (d-6)(d+6)
b). 1-81x^2 = 1^2-(9x)^2 = (1-9x)(1+9x)
c). 16a^2-49c^2 = (4a)^2-(7c)^2 = (4a-7c)(4a+7c)
d). x^10-1 = (x^5)^2-1^2 = (x^5-1)(x^5+1)
e). 16x^2y^2 - 81z^2 = (4xy)^2 - (9z)^2 = (4xy-9z)(4xy+9z)

3).
Try getting into form of ax^2+bx+c

a). Now that we have a >=1 we multiply a with c

a*c = 4*9 = 36

Now we need ax^2+ nx + dx + c so we need to find two numbers n,d that will satisfy these conditions

Condition 1: n*d = a*c
Condition 2: n+d = b

so we need two numbers n,d that add to 12 and multiply to 36. Lets try 6,6

(6)(6) = 36 check
6+6 = 12 = b check

4x^2+6x+6x+9 = 2x(2x+3)+3(2x+3)=(2x+3)(2x+3)

so 4x^2+12x+9 = (2x+3)(2x+3).