If x = 999, y = 1000, z = 1001, then the value of x3+y3+z3-3xyzx-y+z is
A) 1000 B) 9000 C) 1 D) 9
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Answer:
D) 9
Step-by-step explanation:
Because, a3+b3+c3−3abc
= 12(a+b+c)[(a−b)2+(b−c)2+(c−a)2]
Therefore, x3+y3+z3−3xyzx−y+z
= 12(x+y+z)[(x−y)2+(y−z)2+(z−x)2]x−y+z
= 12(999+1000+1001)[1+1+4]999−1000+1001
= 12×6×30001000=9
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