老师，请问x²+［y-（x²）⅓］²=1怎么算？
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(x-y)³+4(y-x)
=(x-y)³-4(x-y)
=(x-y)[(x-y)²-4]
=(x-y)(x-y+2)(x-y-2)
(x-y)²+4(x-y-1)
=(x-y)²+4(x-y)+4
=(x-y+2)²
2(x-1)²-18
=2[(x-1)²-9]
=2(x-1+3)(x-1-3)
=2(x+2)(x-4)
⅓x²+2x-9
=⅓(x²+6x-27)
=⅓(x-3)(x+9)
=(x-y)³-4(x-y)
=(x-y)[(x-y)²-4]
=(x-y)(x-y+2)(x-y-2)
(x-y)²+4(x-y-1)
=(x-y)²+4(x-y)+4
=(x-y+2)²
2(x-1)²-18
=2[(x-1)²-9]
=2(x-1+3)(x-1-3)
=2(x+2)(x-4)
⅓x²+2x-9
=⅓(x²+6x-27)
=⅓(x-3)(x+9)
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