A) 2304 |
B) 2430 |
C) 2034 |
D) 2340 |
C) 2034 |
We know that, \( \Large a^{3} + b^{3} + c^{3} - 3abc \)
= \( \Large \left(a+b+c\right)\frac{1}{2} \left[ \left(a-b\right)^{2} + \left(b-c\right)^{2} + \left(c-a\right)^{2} \right] \)
= \( \Large \left(225 + 226 + 227\right)\frac{1}{2}\left[ 1 + 1 + 4 \right] \)
= \( \Large 678 \times 3 = 2034 \)