(对称多项式）因式分解：x(x+z)^2+y(z+x)^2+z(x+y)^2-4xyz
3个回答
估计是你题打错了,如果是对称多项式,应该是x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz
x(y+z)^2+y(z+x)^2+z(x+y)^2-4xyz
=x(y^2+2yz+z^2)+y(z^2+2xz+x^2)+z(x^2+2xy+y^2)-4xy
=xy^2+xz^2+yz^2+yx^2+zx^2+zy^2+6xyz-4xyz
=xy^2+xz^2+yz^2+yx^2+zx^2+zy^2+2xyz
=yx^2+zx^2+xy^2+xz^2+2xyz+yz^2+zy^2
=(y+z)x^2+(y^2+2yz+z^2)x+yz(y+z)
=(y+z)x^2+(y+z)^2x+yz(y+z)
=(y+z)[x^2+x(y+z)+yz]
=(y+z)(x+y)(x+z)
x^3+3x^2+4是无法分解因式的,可能是符号问题,这有一个
x^3-3x^2+4
=x^3-2x^2-x^2+4
=x^2(x-2)-(x^2-4)
=x^2(x-2)-(x+2)(x-2)
=(x-2)[x^2-(x+2)]
=(x-2)(x^2-x-2)
=(x-2)(x-2)(x+1)
=(x-2)^2(x+1)
x^4+2x^3-11x^2+12x+3b本人实在看不懂3b是什么意思
已知：a,b是实数,且满足a^3-a^2b-b^2+a=0,求证：a=b^2.
题目确实是有问题,将a=b^2代入原式有
b^6-b^5-b^2+b^2=0
即b^6=b^5?
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