without actually calculating , the cubes , find the valuse of
(i) `((1)/(2))^(3)+((1)/(3))^(3)-((5)/(6))^(3)`(ii) `(0.2)^(3)-(0.3)^(3)+(0.1)^(3)`

Explanation

We have
`(i) ((1)/(2))^3+((1)/(3))^3-((5)/(6))^3`
`((1)/(2))^3+((1)/(3))^3+((-5)/(6))^3`
` =a^3+b^3+c^3, " where " a+b+c=(1)/(2)+(1)/(3)+((-5)/(6))=0`
`=3abc " "[because a+b+c=0 rArr a^3+b^3+c^3=3abc]`
` =3xx(1)/(2)xx (1)/(3)xx ((-5))/(6)=(-5)/(12)`
`(ii) (0.2)^3 -(0.3)^3+(0.1)^3`
`=(0.2)^3+(-0.3)63+(0.1)^3`
`a^3+b^3+c^3, " where " a +b+c=0.2 +(-0.3)+0.1=0`
`=3abc " "[because a+b+c=0 rArr a^3+b^3+c^3=3abc]`
`=3xx 0.2xx (-0.3)xx 0.1=-0.018`.