Problem: For integers a,ba, ba,b and ccc, define a,b,c\boxed{a, b, c}a,b,cβ to mean abβbc+caa^{b}-b^{c}+c^{a}abβbc+ca. Then 1,β1,2\boxed{1,-1,2}1,β1,2β equals
Answer Choices:
A. β4-4β4
B. β2-2β2
C. 000
D. 222
E. 444
Solution:
1,β1,2=1β1β(β1)2+21=1β1+2=2\boxed{1,-1,2}=1^{-1}-(-1)^{2}+2^{1}=1-1+2=21,β1,2β=1β1β(β1)2+21=1β1+2=2.