Đặt \(a-b=x\) , \(b-c=y\) và \(c-a=z\)

\(\Rightarrow x+y+z=\left(a-b\right)+\left(b-c\right)+\left(c-a\right)=0\)

Chắc bạn cùng biết  \(x+y+z=0\Rightarrow x^3+y^3+z^3=3xyz\)

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

Chúc bạn học tốt.

\(1,\\ a,=4\left(x-2\right)^2+y\left(x-2\right)=\left(4x-8+y\right)\left(x-2\right)\\ b,=3a^2\left(x-y\right)+ab\left(x-y\right)=a\left(3a+b\right)\left(x-y\right)\\ 2,\\ a,=\left(x-y\right)\left[x\left(x-y\right)^2-y-y^2\right]\\ =\left(x-y\right)\left(x^3-2x^2y+xy^2-y-y^2\right)\\ b,=2ax^2\left(x+3\right)+6a\left(x+3\right)\\ =2a\left(x^2+3\right)\left(x+3\right)\\ 3,\\ a,=xy\left(x-y\right)-3\left(x-y\right)=\left(xy-3\right)\left(x-y\right)\\ b,Sửa:3ax^2+3bx^2+ax+bx+5a+5b\\ =3x^2\left(a+b\right)+x\left(a+b\right)+5\left(a+b\right)\\ =\left(3x^2+x+5\right)\left(a+b\right)\\ 4,\\ A=\left(b+3\right)\left(a-b\right)\\ A=\left(1997+3\right)\left(2003-1997\right)=2000\cdot6=12000\\ 5,\\ a,\Leftrightarrow\left(x-2017\right)\left(8x-2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=2017\\x=\dfrac{1}{4}\end{matrix}\right.\\ b,\Leftrightarrow\left(x-1\right)\left(x^2-16\right)=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=4\\x=-4\end{matrix}\right.\)

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

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

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

\(=\left(a+b\right)\left[b^2\left(1-a^2\right)+a\left(1+a\right)-b.\left(1+a\right)\right]\)

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

Đặt \(\left\{{}\begin{matrix}a+b-c=x\\b+c-a=y\\c+a-b=z\end{matrix}\right.\Leftrightarrow x+y+z=a+b+c\)

Do đó \(A=\left(x+y+z\right)^3-x^3-y^3-z^3\)

\(\Leftrightarrow A=x^3+y^3+z^3+3\left(x+y\right)\left(y+z\right)\left(z+x\right)-x^3-y^3-z^3\\ \Leftrightarrow A=3\left(x+y\right)\left(y+z\right)\left(z+x\right)\)

\(\Leftrightarrow A=3\left(a+b-c+b+c-a\right)\left(b+c-a+c+a-b\right)\left(c+a-b+a+b-c\right)\\ \Leftrightarrow A=3\cdot2b\cdot2c\cdot2a=24abc\)

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

\(2ab^2-a^2b-b^3\)

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

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

( a + b )³ - ( a³ + b³ )

= a³ + 3a²b + 3ab² + b³ - a³ - b³

= 3a²b - 3ab²

= 3ab ( a + b )

\(\left(a+b\right)^3-\left(a^3+b^3\right)\)

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

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

\(=3ab\left(a+b\right)\)

toán lớp 8 nhưg mk mới học lớp 7 thôi!

thông cảm! mik ko thể giúp bn đc

\(2ab^2-a^2b-b^3=-b\left(a^2-2ab+b^2\right)=-b\left(a-b\right)^2.\)

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

`(a+b)^3-3ab(a+b)`

`=(a+b)(a+b)^2-3ab(a+b)`

`=(a+b)[(a+b)^2-3ab]`

`=(a+b)(a^2+2ab+b^2-3ab)`

`=(a+b)(a^2-ab+b^2)`