计算整式(x^2-2xy+4y^2)(x+2y)-8x(x+y)(x-y)(64a^3-27b^3)\(4a-3b)-(
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又做到你的题了

1)(x^2-2xy+4y^2)(x+2y)-8x(x+y)(x-y)

立方和公式:(a+b)(a²-ab+b²)=a³+b³

(x^2-2xy+4y^2)(x+2y)

=(x+2y)[x²-x×2y+(2y)²]

=x³+8y³

8x(x+y)(x-y)

=8x[(x+y)(x-y)]

=8x(x²-y²)

=8x³-8xy²

所以(x^2-2xy+4y^2)(x+2y)-8x(x+y)(x-y)

=x³+8y³-(8x³-8xy²)

= x³+8y³-8x³+8xy²

=-7x³+8xy²+8y³

2)(64a^3-27b^3)(4a-3b)-(4a+3b)^2

立方差公式:a³-b³=(a-b)(a²+ab+b²)

(64a^3-27b^3)(4a-3b)

=[(4a)³-(3b)³]/(4a-3b)

=(4a-3b)[(4a)²+4a×3b+(3b)²]/ (4a-3b)

=[(4a)²+2×4a×3b+(3b)²]-12ab

=(4a+3b)²-12ab

所以(64a^3-27b^3)(4a-3b)-(4a+3b)^2

=(4a+3b)²-12ab-(4a+3b)²

=-12ab

3)(x^4-x+1)/(x^2+2x-3)

题目有没有出错呢?

还是按原来的做

x^4-x+1

=(x^4-2x+1)+x

=[(x^4-2x²+1)+(2x²-2x)]+x

=[(x²-1)²+2x(x-1)]+(x-1)+1

=[(x+1)²(x-1)²+2x(x-1)+(x-1)]+1

=(x-1)[(x+1)²(x-1)+2x+1]+1

=(x-1)(x³-x+x²-1+2x+1)+1

=(x-1)(x³+x²+x)+1

(x^2+2x-3)

=(x+3)(x-1)

所以(x^4-x+1)/(x^2+2x-3)

=【(x-1)(x³+x²+x)+1】/(x+3)(x-1)

=( x³+x²+x )/(x+3)+[1/(x^2+2x-3)]

=(x³+3x²-2x²-6x+7x+21-21) /(x+3)+[1/(x^2+2x-3)]

=x²-2x+7-[21/(x+2)] +[1/(x^2+2x-3)]