解这道英文数学题目find the consecutive number whose cubes（i）differ by

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解这道英文数学题目
find the consecutive number whose cubes（i）differ by 37 (ii)add up to 35

设较小的数是n,
i) 那么：(n+1)^3-n^3=37,
（n+1)^3-n^3=(n+1-n)[(n+1)^2+n(n+1)+n^2]=3n^2+3n+1=37
n^2+n-12=0,(n+4)(n-3)=0,n=-4 或3
所以可以是-4,-3 或3,4.
2）(n+1)^3+n^3=35
（n+1)^3+n^3=2n^3+3n^2+3n+1=35
2n^3+3n^2+3n-34=0
2n^3+3n^2+3n-34=(2n^3-8n^2+8n)+(11n^2-5n-34)
=2n(n-2)^2+(11n+17)(n-2)
=(n-2)(2n^2-4n+11n+17)
=(n-2)(2n^2+7n+17)=0
n=2,（2n^2+7n+17=0无解）
所以是2和3.

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