a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

adwalk · Sep 8, 2019 · a6c54f1 · a6c54f1
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diff --git a/TinyCode2.md b/TinyCode2.md
@@ -582,6 +582,8 @@ print('\n'.join([''.join([('IloveU'[(x-y)%len('IloveU')]if((x*0.05)**2+(y*0.1)**
 42 =(-80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3
  1 = (9t^3 + 1)^3 + (9t^4)^3 + (-9t^4 - 3t)^3 
  2 = (6t^3 + 1)^3 + (-6t^3 + 1)^3 + (-6t^2)^3
+(a+b)^3 - a^3 - b^3 = 3ab(a+b)
+a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)
 ...
 ```
 0和形如9k±4的数不可写成三个立方数之和，其余的都可以。