prove(l.h.s-r.h.s):a^3+b^3+c^3-3abc=(a+b+c) (a^2+b^2+c^2-ab-bc-ca)?

Answer 1

Using the distributive rule on the right, we get
a(a^2 + b^2 + c^2 - ab - bc - ac) +
b(a^2 + b^2 + c^2 - ab - bc - ac) +
c(a^2 + b^2 + c^2 - ab - bc - ac).

Applying it once again, we get:
a^3 + ab^2 + ac^2 - (a^2)b - abc - (a^2)c +
(a^2)b + b^3 + bc^2 - ab^2 - (b^2)c - abc +
(a^2)c + (b^2)c + c^3 - abc - bc^2 - (c^2)a

Combining like terms, this simplifies to:
a^3 + b^3 + c^3 - 3abc

Answer 2

LHS=a^3+b^3+c^3-3abc

is RHS=(a+b+c)^3-(a^2+b^2+c^2-ab-bc-ca) ?

Answer 3

Please leave a few spaces in long formulae so that you don't get troublesome dots.