Question

Evaluate:

(0.2)³ − (0.3)³ + (0.1)³

Answer 1

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.