Question

计算 2xy[4x²y²-3y（xy+x²y）-xy²]

（2x-x²-3）（x³-x²-2）
（x²）³-2x³[x³-x²（4x+1）]
（-1/3xy）²·[xy（2x-y）-3x（xy-y²）]
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Answer 1

2xy [4x²y²-3y(xy+x²y)-xy²]
= 2xy [4x²y²-3xy²-3x²y²-xy²]
= 2xy [x²y²-4xy²]
= 2x³y³-8x²y³

（2x-x²-3）（x³-x²-2）
= (-x²+2x)(x³-x²) - 3(x³-x²) - 2(-x²+2x)+6
= -x³ (x-2)(x-1) - 3x³+3x²+2x²-4x+6
= -x³ (x²-3x+2) - 3x³+3x²+2x²-4x+6
= -x^5+3x^4-2x³ - 3x³+3x²+2x²-4x+6
= -x^5+3x^4-5+5x²-4x+6

（x²）³-2x³[x³-x²（4x+1）]
= x^6-2x³(x³-4x³-x²)
= x^6-2x³(-3x³-x²)
= x^6+ 6x^6+2x^5
= 7x^6+2x^5

（-1/3xy）²·[xy（2x-y）-3x（xy-y²）]
= (1/9x²y²)·(2x²y-xy²-3x²y+3xy²)
= (1/9x²y²)·(-1x²y+2xy²)
= -1/9x^4y³ +2/9x³y^4

Answer 2

=2x³y³-8x²y³