Question

已知a+b=c,求a^3+a^2c+b^2c-abc+b^3的值． (x+1)(x+3)(x+4)(x+6)-280

Answer 1

1.将c=a+b代入得
a^3+a^2*(a+b)+b^2*(a+b)-ab(a+b)+b^3
=2a^3+2b^3
=2(a+b)(a^2-ab+b^2)
这题直接代入就可以算出来了
2.((x+1)(x+6))((x+3)(x+4))-280(首尾一组,中间两个一组)
=(x^2+7x+6)(x^2+7x+12)-280(将x^2+7x看作一整体)
=(x^2+7x)^2+18(x^2+7x)+72-280
=(x^2+7x)^2+18(x^2+7x)-208(用十字相乘法将208分为26和8)
=(x^2+7x+26)(x^2+7x-8)
=(x^2+7x+26)(x+8)(x-1)