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

 Thay a^3+b^3=(a+b)^3 -3ab(a+b) .Ta có : 

Biến đổi vế trái thành: 

a^3+b^3+c^3-3abc 

<=>(a+b)^3 -3ab(a+b) +c^3 - 3abc 

<=>[(a+b)^3 +c^3] -3ab.(a+b+c) 

<=>(a+b+c). [(a+b)^2 -c.(a+b)+c^2] -3ab(a+b+c) 

<=>(a+b+c).(a^2+2ab+b^2-ca-cb+c^2-3ab 

<=>(a+b+c).(a^2+b^2+c^2-ab-bc-ca)

boc vai

(b-c).3+(c-a).3+(a-b).3

=3.(b-c+c-a+a-b)

=3.1

=3

                -Học tốt-

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

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

Cảm ơn Nguyễn Đức Anh :))

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

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

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

Lời giải:

\(a(b^3-c^3)+b(c^3-a^3)+c(a^3-b^3)\)

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

\(=(b^3-c^3)(a-b)-(a^3-b^3)(b-c)\)

\(=(b-c)(a-b)(b^2+bc+c^2)-(a-b)(b-c)(a^2+ab+b^2)\)

\(=(a-b)(b-c)(b^2+bc+c^2-a^2-ab-b^2)\)

\(=(a-b)(b-c)[(c-a)(c+a)+b(c-a)]\)

\(=(a-b)(b-c)(c-a)(c+a+b)\)