第53页

信息发布者:
$6a^3 - 18a^2$
-4
-9
解:​$(1)3xy\ \mathrm {·}2y + x(2x - y^2)$​
​$ = 6xy^2 + 2x^2 - xy^2= 5xy^2 + 2x^2$​
​$ (2)(2\ \mathrm {m} + n)(\mathrm {m^2} - n)$​
​$ = 2\ \mathrm {m} ·\mathrm {m^2} + 2\ \mathrm {m} ·(-n) + n ·\mathrm {m^2} + n ·(-n)$​
​$ = 2\ \mathrm {m^3} - 2\ \mathrm {m}n +\mathrm {m^2}n - n^2$​
​$ = 2\ \mathrm {m^3} +\mathrm {m^2}n - 2\ \mathrm {m}n - n^2$​
解:​$(1)(x^2 + ax - 2)(3x + b)$​
​$ = x^2\ \mathrm {·}3x + x^2\ \mathrm {·}b + ax\ \mathrm {·}3x + ax\ \mathrm {·}b - 2\ \mathrm {·}3x - 2\ \mathrm {·}b$​
​$ = 3x^3 + bx^2 + 3ax^2 + abx - 6x - 2b$​
​$ = 3x^3 + (b + 3a)x^2 + (ab - 6)x - 2b$​
∵展开式中不含​$x$​的一次项,常数项是​$-6$​
∴​$ab - 6 = 0,$​​$-2b = -6,$​解得​$b = 3,$​​$a = 2$​
​$ (2)(a + b)(a^2 - ab + b^2)$​
​$ = a ·a^2 - a ·ab + a ·b^2 + b ·a^2 - b ·ab + b ·b^2$​
​$ = a^3 - a^2b + ab^2 + a^2b - ab^2 + b^3$​
​$ = a^3 + b^3$​
​$ $​当​$a = 2,$​​$b = 3$​时,原式​$= 2^3 + 3^3 = 35$​