Question

[Kiều Ngọc Tú Anh]

cho x+ y + z = 0 cmr x⁷+y⁷+z⁷=7xyz(x²y²+y²z²+z²x²)

Answer 1

[Phong Thần]

Ta có:

\(x+y+z=0\)

\(\Rightarrow x+y=-z\)

Ta lại có:

\(x^7+y^7\)

\(=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^4y^x-x^3y^4\)

\(=\left(x^3+y^3\right)\left(x^4+y^4\right)-x^3y^3\left(x+y\right)\)

\(=\left(x^3+y^3\right)\left(x^4+y^4\right)+x^3y^3z\) ( Thay x + y = -z )

Ta sẽ đi tính \(x^3+y^4;x^4+y^4\)
Lại có​:

1/ \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)=-z^3+3xyz\)

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

\(\Rightarrow x^4+y^4=\left(x^2+y^2\right)^2-2x^2y^2=\left(z^2-2xy\right)^2-2x^2y^2=z^4-4xyz^2+2x^2y^2\)

Như vậy \(x^7+y^7=\left(-z^3+3xyz\right)\left(z^4-4xyz^2+2x^2y^2\right)+x^3y^3z\)

\(\Rightarrow x^7+y^7=-z^7+7xyz^5-14x^2y^2z^3+7x^3y^3z\)

\(\Rightarrow x^7+y^7+z^7=7xyz^5-14x^2y^2z^3+7x^3y^3z\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left(z^4-2xyz^2+x^2y^2\right)\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left[z^2\left(z^2-2xy\right)+x^2y^2\right]\)

Mà \(z^2-2xy=x^2+y^2\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left[z^2\left(x^2+y^2\right)+x^2y^2\right]\)

\(\Rightarrow x^7+y^7+z^7=7xyz\left(x^2z^2+y^2z^2+x^2y^2\right)\)