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The value of [(0.111)3+(0.222)3(0.333)3+(0.333)2×(0.222)]2=[(0.111)^3+(0.222)^3-(0.333)^3+(0.333)^2\times(0.222)]^2=

A0

B1

C2

D3

Answer:

A. 0

Read Explanation:

If a3+b3+c3=3abca^3 +b^3 +c^3 = 3abc

then, a+b+c=0a+b+c=0

a = 0.111, b = 0.222, c = -0.333

(0.111)3+(0.222)3+(0.333)30.333×0.333×0.222(0.111)^3+(0.222)^3+(-0.333)^3-0.333\times0.333\times0.222

=(0.111)3+(0.222)3+(0.333)33×0.111×0.333×0.222=(0.111)^3+(0.222)^3+(-0.333)^3-3\times0.111\times0.333\times0.222

    0.111+0.2220.333=0\implies0.111+0.222-0.333=0


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