Problem: What is the smallest integer larger than (3β+2β)6?
Answer Choices:
A. 972
B. 971
C. 970
D. 969
E. 968
Solution:
Since a6+b6=(a2+b2)(a4βa2b2+b4) and 3ββ2β<1,
(3β+2β)6+(3ββ2β)6=(10)(97)=970
and 970 is the smallest integer larger than (3β+2β)6.