Question

求这些数学题的答案

求这些数学题的答案需要过程，谢谢

Answer 1

(7)原式=(a-b)³ + 4ab(a-b)
=(a-b)[(a-b)² + 4ab]
=(a-b)(a² - 2ab + b² + 4ab)
=(a-b)(a+b)²

追答

(2)原式=3(a³+b³+c³-3abc)
=3(a³+3a²b+3ab²+b³+c³-3abc-3a²b-3ab²)
=3[(a+b)³+c³ - (3abc+3a²b+3ab²)]
=3[(a+b)³+c³ - 3ab(c+a+b)]
=3(a+b+c)[(a+b)²-(a+b)c+c² - 3ab]
=3(a+b+c)(a²+b²+c²-ab-ac-bc)

5、a³+b³=(a+b)(a²-ab+b²)
=6(a²-ab+b²)=72
∴a²-ab+b²=12①
(a+b)²=a²+2ab+b²=36②
①×2 + ②：3a² + 3b²=60
∴a²+b²=20
6、两式相减：x³-x-y³+y=0
(x³-y³) - (x-y)=0
(x-y)(x²+xy+y²) - (x-y)=0
(x-y)(x²+xy+y²-1)=0
∵x≠y
∴x²+xy+y²-1=0
则x²+xy+y²=1

7、原式=x^12•(x³+x²+x+1) + x^8•(x³+x²+x+1) + x^4•(x³+x²+x+1) + (x³+x²+x+1)
=(x³+x²+x+1)(x^12 + x^8 + x^4 + 1)
=[x²(x+1)+(x+1)][x^8(x^4 + 1) + (x^4 + 1)]
=(x+1)(x²+1)(x^4 + 1)(x^8 + 1)