shok

multumesc gaby pentru superscript

bazil

Soluţia finală ar fi: 3*(x+y)(y+z)(z+x).
Poate mâine am timp să o transpun!

bazil

(x+y+z)^3 - x^3 - y^3 - z^3 =
[(x+y+z)^3 - x^3] - (y^3 + z^3) =
(x+y+z-x) [(x+y+z)^2 + x*(x+y+z) +x^2] - (y+z) (y^2 - yz + z^2) =
(y+z) (x^2 + y^2 + z^2 + 2xy + 2xz + 2yz + x^2 + xy + xz + x^2) - (y+z) (y^2 - yz + z^2)=
(y+z) (3*x^2 + y^2 + z^2 + 3xy + 3xz + 2yz) - (y+z) (y^2 - yz + z^2)=
(y+z) (3*x^2 + y^2 + z^2 + 3xy + 3xz + 2yz - y^2 + yz - z^2)=
(y+z) (3*x^2 + 3xy + 3xz + 3yz)=
(y+z) [3x (x+y) + 3z (x+y)]=
3 (y+z) (x+y) (x+z).

d-attila

The same solution like the previous, just on a different (more detailed) way:
a+b+c)^3 –( a^3 + b^3 + c^3) =
(a+b+c)(a+b+c)(a+b+c) –( a^3 + b^3 + c^3)  =
(a^2 + b^2 + c^2 + 2ab + 2ac + 2bc)(a+b+c) –( a^3 + b^3 + c^3) =
a^3 + b^3 + c^3 + 3a^2b + 3ab^2 + 3a^2c + 3ac^2 + 3b^2c + 3bc^2 + 6abc–( a^3 + b^3 + c^3)=
3a^2b + 3ab^2 + 3a^2c + 3ac^2 + 3b^2c + 3bc^2 + 6abc=
3[a^2b + ab^2 + a^2c + ac^2 + b^2c + bc^2 + 2abc] =
3[a^2b + ab^2 + a^2c + abc + ac^2 + bc^2+ b^2c+abc] =
3[ab(a+b)+ac(a+b)+c^2(a+b)+bc(a+b)]=
3(a+b)(ab+ac+c^2+bc)=
3(a+b)[a(b+c)+c(b+c)]=
3(a+b)(a+c)(b+c)
Sorry for replacing for a,b,c,....