\(\left(\left(x+y\right)^2\right)^3+\left(\left(x-y\right)^2\right)^3\)

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

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

\(=2\left(x^2+y^2\right)\left(\left(x+y\right)^4-\left(x^2-y^2\right)^2+\left(x-y\right)^4\right)⋮\left(x^2+y^2\right)\)

\(\left(x+y\right)^6+\left(x-y\right)^6\)

\(=\left[\left(x+y\right)^2\right]^3+\left[\left(x-y\right)^2\right]^3\)

\(=\left[\left(x+y\right)^2+\left(x-y\right)^2\right]\left(...\right)\)

\(=\left(x^2+2xy+y^2+x^2-2xy+y^2\right)\left(...\right)\)

\(=\left(2x^2+2y^2\right)\left(...\right)\)

\(=2\left(x^2+y^2\right)\left(...\right)⋮x^2+y^2\left(đpcm\right)\)

1) A=\(\left(x+y\right)^6+\left(x-y\right)^6=\left[\left(x+y\right)^2+\left(x-y\right)^2\right]\left[binh-phuong-thieu\right]\)

                                             \(=2\left(x^2+y^2\right)\left[binh-phuong-thieu..\right]\)=> A chia hết cho x²+y²

2)  gọi dư của phép chia là ax+b

 ta có f(1) = a+b =51

         f(-1) = -a+b =1 

=> b =26 ; a =25

Vậy dư là : 25x + 26

\(\left(x+y+z\right)⋮6\Rightarrow\left(x+y+z\right)⋮2\)

x, y, z không thể đồng thời cả 3 số cùng lẻ ; nghĩa là phải có 1 số chẵn

\(\left\{{}\begin{matrix}\left(x.y.z\right)⋮2\Rightarrow3\left(xyz\right)⋮6\\\left(\left(x-y\right)\left(x+y\right)\left(x+y+z\right)\right)⋮6\end{matrix}\right.\)

\(\Rightarrow A⋮6\Rightarrow dpcm\)