A) 0
B) 343
C) 392
D) 2863
Correct Answer: A
Solution :
[a] Here, \[(x-\,y)=7\] Then, \[\therefore \,\,\,{{a}^{3}}-\,{{b}^{3}}=(a-\,b)\,\,({{a}^{2}}+ab+{{b}^{2}})\] \[{{(x-\,15)}^{3}}-\,{{(y-\,8)}^{3}}=?\] \[\Rightarrow (x-15-y+8)\,\,[{{(x-15)}^{2}}+(x-15)\,\,(y-8)+{{(y-8)}^{2}}]\]\[\Rightarrow (x-y-7)\,\,[{{(x-15)}^{2}}+(x-15)\,\,(y-8)+{{(y-8)}^{2}}]\]\[\Rightarrow (7-7)\,\,[{{(x-15)}^{2}}+(x-15)\,\,(y-8)+{{(y-8)}^{2}}]\] \[\Rightarrow 0\times [{{(x-15)}^{2}}+(x-15)\,\,(y-8)+{{(y-8)}^{2}}]\] \[\Rightarrow 0\]
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