The value of [(0.111)3+(0.222)3−(0.333)3+(0.333)2×(0.222)]2=
A0
B1
C2
D3
Answer:
A. 0
Read Explanation:
If a3+b3+c3=3abc
then, a+b+c=0
a = 0.111, b = 0.222, c = -0.333
(0.111)3+(0.222)3+(−0.333)3−0.333×0.333×0.222
=(0.111)3+(0.222)3+(−0.333)3−3×0.111×0.333×0.222
⟹0.111+0.222−0.333=0