Show that $(a+b+c)^3 = a^3 + b^3 + c^3+ (a+b+c)(ab+ac+bc)$ As stated in the title, I'm supposed to show that $(a+b+c)^3 = a^3 + b^3 + c^3 + (a+b+c)(ab+ac+bc)$. 
My reasoning: 
$$(a + b + c)^3 = [(a + b) + c]^3 = (a + b)^3 + 3(a + b)^2c + 3(a + b)c^2 + c^3$$
$$(a + b + c)^3 = (a^3 + 3a^2b + 3ab^2 + b^3) + 3(a^2 + 2ab + b^2)c + 3(a + b)c^2+ c^3$$ 
$$(a + b + c)^3 = a^3 + b^3 + c^3 + 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + (3a^2b + 3a^2c + 3abc) + (3ab^2 + 3b^2c + 3abc) + (3ac^2 + 3bc^2 + 3abc) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3a(ab + ac + bc) + 3b(ab + bc + ac) + 3c(ac + bc + ab) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3(a + b + c)(ab + ac + bc) - 3abc$$
$$(a + b + c)^3 = (a^3 + b^3 + c^3) + 3[(a + b + c)(ab + ac + bc) - abc]$$
It doesn't look like I made careless mistakes, so I'm wondering if the statement asked is correct at all. 

$(a+b+c)^3=a^3+b^3+c^3+3(a+b)(b+c)(a+c)$ with $3(a+b)(b+c)(c+a) = 3a^2b + 3a^2c + 3ab^2 + 3b^2c + 3ac^2 + 3bc^2 + 6abc$
I guess the right factorization
