Question

2. If a = 0,b= -1, find the value of
(a) a³+b³+3a²b+3ab²
(b) a²+2ab+b²
(c) a²–b²
(d) [a+b] [a–b]
(e) [a+b]³
(f) [a+b]²​

Answer 1

Answer:

1. (0³)+(-1)³+3(0)(-1)+3(0)(-1)²

= 0-1-3+3

= -1

2. 0+2(0)(-1)+(-1)²

= 1

Answer 2

Answer:

a) 0+(-1)³+3(0)²(-1)+3(0)(-1)² =0-1+0+0= -1

b) 0²+2(0)(-1)+(-1)² = 0+0+1=1

c) 0²-(-1)² = 0-1 = -1

d) (a+b)(a-b) = a²- b² = -1 ( from c)

e) (a+b)³ = a³+b³+3a²b+3ab² = -1 (from a)

f) (a+b)² = a²+2ab+b² = 1 (from b)