Solution:
Given:
a = 299
b = 298
c = 297
Formula used:
[a3 + b3 + c3 – 3abc] = 21× (a + b + c)[(a – b)2 + (b – c)2 + (c – a)2]
Calculation:
2a3 + 2b3 + 2c3 – 6abc
= 2[a3 + b3 + c3 – 3abc]
= 2×21× (a + b + c)[(a – b)2 + (b – c)2 + (c – a)2]
= (299 + 298 + 297) [(299 – 298)2 + (298 – 297)2 + (297 – 299)2]
= 894 [12 + 12 + 22]
= 894×6
= 5364