若a-b=-3,ab=4,求⑴（a+b）² ⑵a³-b³?

已知
{ a - b = -3
{ ab = 4

那麼
(a - b)² = (-3)²
a² - 2ab + b² = 9
a² - 2(4) + b² = 9
a² - 8 + b² = 9
a² + b² = 17

(1)
(a + b)²
= a² + 2ab + b²
= 17 + 2(4)
= 17 + 8
= 25

(2)
a³ - b³
= (a - b)(a² + ab + b²)
= (-3)(17 + 4)
= (-3)(21)
= -63

1. 25
2. -63

(1)
a-b=-3
(a-b)^2=9
a^2 -2ab+b^2=9
a^2+2ab+b^2-4ab=9
(a+b)^2=9+4(4)
(a+b)^2=25

(2)
a-b=-3
(a-b)^3=-27
a^3+(3)(a^2)(-b)+(3)(a)(-b)^2-b^3=-27
a^3-b^3=-27+3(4a)-3(4b)
a^3-b^3=-27+(12)(a-b)
a^3-b^3=-27+12(-3)
a^3-b^3=-63