Without calculating the actual cube find the value of `(-12)^3+7^3+5^3`and `28^3+(-15)^3+(-13)^3`

Explanation

we know that
`a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)`

(i)`(-12)^3+(7)^3+(5)^3=3xx(-12)xx7xx5+(-12+7+5)xx((-12)^2+7^2+5^2-(-12)7-7xx5-(-12)5)`
`=(-36)(35)+0`
`=-1260`
(ii)`(28)^3+(-15)^3+(-13)^3=3xx28xx(-15)xx(-13)+(28-13-15)((28)^2+(-15)^2+(-13)^2-28(-15)-(-15)(-13)-28(-13)`
`=84xx195+0`
`=16380`