Question

Chứng minh rằng đa thức \(\left(x+y\right)^6+\left(x-y\right)^6\) chia hết cho đa thức \(x^2+y^2\)

Answer 1

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

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

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

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

Answer 2

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

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

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

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

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

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