we realize that
(a+b+c)^3-a^3 has a factor (b+c) and b^3 + c^3 has a factor b+c
so we proceed by combining
(a+b+c) ^3- a^3-b^3-c^3
= ((a+b+c) ^3- a^3)-(b^3+c^3)
= (b+c)((a+b+c)^2 + a(a+b+c) + a^2) - (b+c)(b^2+c^2-bc)
= (b+c)((a+b+c)^2+ a(a+b+c) + a^2 - b^2 - c^2 + bc)
= (b+c)(a^2+b^2+c^2+2ab + 2ac + 2bc a^2+ab+ac + a^2 - b^2 - c^2 + bc)
= (b+c)(3a^2 + 3ab + 3bc + 3ca)
= 3(b+c)(a^2+ab+bc+ca)
= 3(b+c)(a+c)(a+b)
4 comments:
Thanks for wonderful pages.
I also like this because it helps explain "why" 3^3 + 4^3 + 5^3 = 6^3
I can do this better)if you take a=-b,and you will see (-b+b+c)^3-(-b)^3-b^3-c^3=c^3-c^3=0,so (a+b+c)^3-a^3-b^3-c^3 shares on (a+b) Bezout 's theorem.then take b=-c and also sheres on b+c. Then take c=-a and also sheres on c+a.(a+b+c)^3-a^3-b^3-c^3 =k(a+b)(b+c)(c+a).
K is unknown, so find her
If a=1,b=1,c=0》(1+1+0)^3-1^3-1^3-0=k (1+1)(1+0)(0+1)
8-2=k*2
K=3
(a+b+c)^3-a^3-b^3-c^3 =3 (a+b)(b+c)(c+a)
See this one-
((a+b)+c)^3-a^3-b^3-c^3
(a+b)^3+c^3+3(a+b)(c)(a+b+c)-a^3-b^3-c^3
a^3+b^3+3ab(a+b)+3(a+b)(c)(a+b+c)-a^3-b^3
3(a+b)(ab+c(a+b+c))
3(a+b)(ab+ac+bc+c^2)
3(a+b)(a(b+c)+c(b+c))
3(a+b)(b+c)(c+a)
Excellent
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