Without actually calculating the cubes, find the value of each of the following:

 (i) (−12)3+(7)3+(5)3

(ii) (28)3+(−15)3+(−13)3

Solution:

(i) (−12)3+(7)3+(5)3

Let a = −12

b = 7

c = 5

We know that if x+y+z = 0, then x3+y3+z3=3xyz.

Here, −12+7+5=0

(−12)3+(7)3+(5)3 = 3xyz

= 3×-12×7×5

= -1260

(ii) (28)3+(−15)3+(−13)3

Solution:

(28)3+(−15)3+(−13)3

Let a = 28

b = −15

c = −13

We know that if x+y+z = 0, then x3+y3+z3 = 3xyz.

Here, x+y+z = 28–15–13 = 0

(28)3+(−15)3+(−13)3 = 3xyz

= 0+3(28)(−15)(−13)

= 16380