若a/(X^2-YZ)=b/(Y^2-ZX)=c/(Z^2-XY)求证：aX+bY+cZ=(X+Y+Z)(a+b+c)

Question

若a/(X^2-YZ)=b/(Y^2-ZX)=c/(Z^2-XY)
求证：aX+bY+cZ=(X+Y+Z)(a+b+c)

Answer 1

证明：首先,令a/(X^2-YZ)=b/(Y^2-ZX)=c/(Z^2-XY) = m （用换元法）
所以,上式变形为
a = m(X^2-YZ)
b = m(Y^2-ZX)
c = m(Z^2-XY)
所以,aX+bY+cZ = m(X^2-YZ)*X + m(Y^2-ZX)*Y + m(Z^2-XY)*Z
= m(X^3+Y^3+Z^3 - 3XYZ)
= m [X^3+(X^2)Y+(X^2)Z + Y^3+(Y^2)X+(Y^2)Z + Z^3+(Z^2)X+(Z^2)Y
-(X^2)Y-(X^2)Z -(Y^2)X-(Y^2)Z -(Z^2)X-(Z^2)Y -3XYZ]
= m [X^3+(X^2)Y+(X^2)Z + Y^3+(Y^2)X+(Y^2)Z + Z^3+(Z^2)X+(Z^2)Y
-(X^2)Y-XY^2-XYZ -(Y^2)Z-YZ^2-XYZ -(Z^2)X-ZX^2-XYZ]
= m [X^2(X+Y+Z) + Y^2(X+Y+Z) + Z^2(X+Y+Z) - XY(X+Y+Z) - YZ(X+Y+Z) - ZX(X+Y+Z)]
= (X+Y+Z)[m(X^2-YZ) + m(Y^2-ZX) + m(Z^2-XY)]
=(X+Y+Z)(a+b+c)
即,aX+bY+cZ=(X+Y+Z)(a+b+c)
（证毕）