Question

Докажите тождество: а(в+с)^2 + b(c+a)^2 + c(a+b)^2 - 4abc = (a+b) (b+c)(c+a)

Answer 1

$a(b+c)^2+b(c+a)^2+c(a+b)^2-4abc= \\ =a(b+c)^2+(b(a^2+2ac+c^2)+c(a^2+2ab+b^22)-4abc)= \\ =a(b+c)^2+(a^2b+bc^2+a^2c+b^2c)= \\ =a(b+c)^2+(a^2+bc)(b+c)= \\ =(a(b+c)+(a^2+bc))(b+c)=((a+b)c+a(a+b))(b+c)= \\ =\boxed{(a+b)(c+a)(b+c)}$