Factorization a (1-B) ^ 2-1 + 2b-b ^ 2

Factorization a (1-B) ^ 2-1 + 2b-b ^ 2

Factorization: (C-A) 2-4 (B-C) (a-b)

（c-a）2-4（b-c）（a-b）=c2-2ac+a2-4ab-4bc+4ac+4b2=（c+a）2-4b（c+a）+4b2=（c+a-2b）2．

Factorization of (A-C) ^ 2 - (B-D) ^ 2

Square difference
Original formula = (A-C + B-D) (A-C-B + D)

Factorization of (a + B-C) ^ 3 - (a ^ 3 + B ^ 3 + C ^ 3)

Original formula = (a + b) ^ 3-3 (a + b) ^ 2 * C + 3 (a + b) * C ^ 2-C ^ 3 - (a ^ 3 + B ^ 3 + C ^ 3)
=(a^3+b^3+3a^2b+3ab^2)-3(a^2+2ab+b^2)*c+3ac^2+3bc^2-c^3-(a^3+b^3+c^3)
=a^3+3a^2b+3ab^2+b^3-3a^2c-6abc-3b^2c+3ac^2+3bc^2-a^3-b^3-2c^3
=3a^2b+3ab^2-3a^2c-6abc-3b^2c+3ac^2+3bc^2-2c^3
=3a^2(b-c)+3a(b^2-2bc+c^2)-3b^2c+3bc^2-2c^3
=3a^2(b-c)+3a(b-c)^2-3bc(b-c)-2c^3
=3(b-c)(a^2+ab-ac-bc)-2c^3
It's not easy to change in the back
If the coefficient of C in the preceding bracket is positive, the result of this problem is
3(a+b)(a+c)(b+c)
There may be some problems with the title

Factorization (B-C) (B + C) ^ 3 + (C-A) (c + a) ^ 3 + (a-b) (a + b) ^ 3

(b-c)(b+c)³+(c-a)(c+a)³+(a-b)(a+b)³
=(b²-c²)(b²+2bc+c²)+(c²－a²)(c²＋2ac+c²)+(a²－b²)(a²+2ab+b²)
=2b³c-2bc³+2c³a﹣2ca³+2a³b-2ab³
=2(b³c-bc³+c³a-ca³+a³b-ab³)
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Factorization a ^ 2-B ^ 2 + C ^ 2
(^ 2 for square)

Factorization of x ^ 2 + (a + B + C) x + (a + b) C
Multiplication by cross

1 a+b
1 c
Original formula = (x + A + b) (x + C)

The square factorization of a's square B + 4AB + 4-b

a²b²+4ab+4-b²
=(ab)²+4ab+4-b²
=(ab+2)²-b²
=(ab+b+2)(ab-b+2)
If you still have any questions, please ask!

1, the result of the square of X * x + X * 1 / 2-1 / 2x-1 / 2x must be 0, because? 2, given the algebraic formula - x + 3Y = 9, then 2x-6y + 19 is? 1?

-x+3y = 9,2x-6y + 19= -2(x+3y) + 19 = -18 + 19 = 1

If 3axb and 2ab1-y are similar, then the value of x2011-y2010 is ()
A. 1B. -3C. -1D. 0

∵ the binomial 3axb and 2ab1-y are of the same kind, ∵ x = 1, y = 0, ∵ x2011-y2010 = 1-0 = 1