Question

cmr

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

(a+b)(b+c)(c+a)+4abc=c(a+b)^2+a(b+c)^2+b(c+a)^2

Answer 1

2(a^3+b^3+c^3-3abc)

=2[(a+b)^3+c^3-3ab(a+b)-3abc]

=2[(a+b+c)(a^2+b^2+c^2-ab-bc-ac)]

=(a+b+c)[(a-b)^2+(b-c)^2+(a-c)^2]

c(a+b)^2+a(b+c)^2+b(c+a)^2

=ca^2+cb^2+2abc+ab^2+2abc+ac^2+bc^2+2abc+ba^2

=4abc+a^2c+a^2b+c^2b+c^2a+b^2a+b^2c+2abc

=4abc+(a+b)(b+c)(a+c)