Factor (a + b + c)^3 - a^3 - b^3 - c^3?

Question

Factor (a + b + c)^3 - a^3 - b^3 - c^3.


After expanding on paper, I ended up with the following:

3(a^2 b + a^2 c + ab^2 + 2abc+a c^2 + b^2 c + bc^2).

I then used the famous wolfram site to further simplify the crazy polynomial in the parentheses.

It turns out that

(a^2 b + a^2 c + ab^2 + 2abc+a c^2 + b^2 c + bc^2) can be simplified to look like this:

(a + b)(a + c)(b + c).

So, the final answer is

3(a + b)(a + c)(b + c).

Question:

How is (a^2 b + a^2 c + ab^2 + 2abc+a c^2 + b^2 c +
bc^2) simplified or factored to be (a + b)(a + c)(b + c) by hand or without using wolfram?

Answer 1

Question :

How is (a^2 b + a^2 c + ab^2 + 2abc+a c^2 + b^2 c + bc^2)

simplified or factored to be (a + b)(a + c)(b + c) by hand or without using wolfram?



Answer :

a^2*b + a^2*c + a*b^2 + 2abc + a*c^2 + b^2*c + b*c^2

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

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

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

= (a + c)[a(b + c) + b(b + c)]

= (a + c)(a + b)(b + c)

= (a + b)(a + c)(b + c)

Answer 2

Regard it as a polynomial in say a:



p(a) = (a + b + c)^3 - a^3 - b^3 - c^3



Then p(-b) = c^3 + b^3 - b^3 - c^3 = 0.

This means (a+b) is a factor of p(a) by the factor theorem.



Similarly,

p(-c) = b^3 + c^3 - b^3 - c^3 = 0.

This means (a+c) is a factor of p(a) by the factor theorem.



Since the given polynomial is symmetric in a,b,c, i.e. any permutation of a,b,c gives back the same polynomial, the fact that (a+b) and (a+c) are factors implies that (b+c) is also a factor. This could also be seen, by regarding the original polynomial as a polynomial in b (or in c).



Conclusion:



(a + b + c)^3 - a^3 - b^3 - c^3 = (a+b)(a+c)(b+c).

Answer 3

a^2 b + a^2 c + ab^2 + 2abc+a c^2 + b^2 c + bc^2



This can be solved with a little grouping. First, factor out the a terms:

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



Simplify:

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



Factor out b+c:

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



Factor:

(b + c) (a + b) (a + c)