Factorization 2 (7 19:48:9) 1. Given A-B = 5, ab = 3, find the value of the algebraic formula A ^ 3b-2a ^ 2B ^ 2 + AB ^ 3   2. First simplify, in the evaluation: [(3a-7) ^ 2 - (a + 5) ^ 2] / (4a-24), where a = 1 / 50   3. Factorization: (x ^ 2-2x) (x ^ 2-2x + 2) + 1

Factorization 2 (7 19:48:9) 1. Given A-B = 5, ab = 3, find the value of the algebraic formula A ^ 3b-2a ^ 2B ^ 2 + AB ^ 3   2. First simplify, in the evaluation: [(3a-7) ^ 2 - (a + 5) ^ 2] / (4a-24), where a = 1 / 50   3. Factorization: (x ^ 2-2x) (x ^ 2-2x + 2) + 1

1. The original formula = AB (a ^ 2-2ab + B ^ 2) = AB (a-b) ^ 2
Substituting A-B = 5 and ab = 3
Then the original formula = 3 * 5 ^ 2 = 75
2. & shy; original formula = (3a-7-a-5) (3a-7 + A + 5) / (4 (a-6))
=(2a-12)(4a-2)/(4(a-6))
=4(a-6)(2a-1)/(4(a-6))
=2a-1
When a = 1 / 50
Original formula = (1 / 25) - 1
=-0.75
3. Let x ^ 2-2x = t
Original formula = t (T + 2) + 1
=t^2+2t+1
=(t+1)^2
=(x^2-2x+1)^2
=(x-1)^4

Solving calculation problems with factorization (31 18:26:5)
2005²—2004×2006=              

2005²—2004×2006
＝2005^2-(2005-1)(2005+1)
=2005^2-2005^2+1
=1

Factorization (23 11:35:5)
Decomposition factor: (x + y) ^ 2-4 (x + y) m + 4m ^ 2

（X+Y）^2-4(X+Y)M+4M^2
=(x+y-2m)²
[take x + y as a, 2m as B, that is a & sup2; - 2Ab + B & sup2; = (a-b) & sup2;]

How to get x + 1 / x = 4 from x2-4x + 1 = 0?
 

Because x = 0 is not the solution of the original equation
So both sides of the equation can be divided by X
We get x-4 + 1 / x = 0
Namely
x+1/x=4