Question

without actually calculating the cubes,find the following value.(0. 2)^3-(0. 3)^3+(0. 1)^3

Answer 1

hey mate


using identity:
x3 + y3 + z3  - 3xyz = (x + y + z)(x2 + y2 + z2 – xy – yz – zx)

if x + y + z = 0,
then, x3 + y3 + z3  - 3xyz = 0
or x3 + y3 + z3  = 3xyz

(0.2)3 - (0.3)3 + (0.1)3 can be written as (0.2)3+ (-0.3)3 + (0.1)3    //since (-x)3 = -x3 //
here, (0.2) + (-0.3) + (0.1) = 0.3 - 0.3 = 0
so, (0.2)3 - (0.3)3 + (0.1)3 = 3*(0.2)(-0.3)(0.1) = - 0.018

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Answer 2

acc to an equation
if
a+b+c=0 them
a^3 + b^3 +c^3 =3abc
so here
3abc= 3(2×-3×1)÷1000
= -0.018