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Question
Solution
Given (0.2)3 − (0.3)3 + (0.1)3
We shall use the identity `a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2 +b^2 + c^2 - ab -bc-ca)`
Let Take a = 0.2,b=0.3 ,c=0.1
`a^3 + b^3 +c^3 - 3abc = (a+b+c)(a^2+b^2 +c^2 - ab-bc - ca)`
`a^3 + b^3 +c^3 = (a+b+c)(a^2+b^2 +c^2 - ab-bc - ca)+3abc`
\[a^3 + b^3 + c^3 = \left( 0 . 2 - 0 . 3 + 0 . 1 \right)\left( a^2 + b^2 + c^2 - ab - bc - ca \right) + 3abc\]
\[a^3 + b^3 + c^3 = 0 \times \left( a^2 + b^2 + c^2 - ab - bc - ca \right) + 3abc\]
`a^3+b^3+c^3 = +3abc`
`(0.2)^3 - (0.3)^3 + (0.1)^3 = 3 xx 0.2 xx 0.3 xx 0.1`
` = -0.018`
Hence the value of (0.2)3 − (0.3)3 + (0.1)3 is -0.018.
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