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