\(3\left(a+3b\right)\left(b+3c\right)\left(c+3a\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

\(=\left(a+b-2c+b+c-2a\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]+\left(c+a-2b\right)^3\)

\(=\left(c+a-2b\right)^3-\left(a-2b+c\right)\left[\left(a+b-2c\right)^2-\left(a+b-2c\right)\left(b+c-2a\right)+\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(c+a-2b\right)^2-\left(a+b-2c\right)^2+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(c+a-2b+a+b-2c\right)\left(c+a-2b-a-b+2c\right)+\left(a+b-2c\right)\left(b+c-2a\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b\right)-\left(a+b-2c\right)\left(2a-b-c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(3c-3b-a-b+2c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(2a-b-c\right)\left(5c-a-4b\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left[\left(b+c-2a\right)\left(a+4b-5c\right)-\left(b+c-2a\right)^2\right]\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(a+4b-5c-b-c+2a\right)\)

\(=\left(c+a-2b\right)\left(b+c-2a\right)\left(3a+3b-6c\right)\)

\(=3\left(c+a-2b\right)\left(b+c-2a\right)\left(a+b-2c\right)\)

\(B=\left(a+b-2c\right)^3+\left(b+c-2a\right)^3+\left(c+a-2b\right)^3\)

Đặt: \(a+b-2c=x;b+c-2a=y;c+a-2b=z\)

\(\Rightarrow B=x^3+y^3+z^3=\left(x+y+z\right)^3-3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

Ta thấy: \(x+y+z=a+b-2c+b+c-2a+c+a-2b=0\)

\(x+y=a+b-2c+b+c-2a=2b-a-c\)

\(y+z=b+c-2a+c+a-2b=2c-a-b\)

\(z+x=c+a-2b+a+b-2c=2a-b-c\)

Thay vào B \(\Rightarrow B=0-3\left(2b-a-c\right)\left(2c-a-b\right)\left(2a-b-c\right)\)

Vậy \(B=-3\left(2b-a-a\right)\left(2c-a-b\right)\left(2a-b-c\right).\)

a,x⁴+2x²+5x³+10x-2x² - 4

= x².(x² +2) + 5 - x.(x²+2) - 2.(x²+2)

= (x² + 5x -2).(x²+2)

b,

Chúc bạn học tốt
Bình chọn cho mình nhé <3

a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)

b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)

c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)

\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)

\(=3b\left(2a-b\right)\)

`a, 4x^2-1 = (2x+1)(2x-1)`

`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`

`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`