Question

evaluate (0.99)^3
using algebraic identity and manually

Answer 1

(1-0.01)^3
apply (a-b)^3=a^3 +b^3+ 3ab^2-3a^2b
here a= 1 and b = 0.01
1+0.000001+3x1x0.01x0.01 - 3x1x1 x0.01
1+0.000001+0.0003-0.03 = 0.970301

Answer 2

(1-0.01)^3

Using (a-b)^3=a^3 +b^3+ 3ab^2-3a^2b

a= 1 and b = 0.01

1+0.000001 + 3x1x0.01 x 0.01-3x1x1 x0.01

1+0.000001+0.0003-0.03 =

0.970301

Ans 0.970301