How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$? How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?
 A: You just need to do a little rearrangement :
$$a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$$
         $$a^2b+a^2c+b^2c+bc^2+ab^2+abc+abc+ac^2$$
$$a^2(b+c)+bc(b+c)+ab(b+c)+ac(b+c)$$
$$(b+c)(a^2+bc+ab+ac)$$
$$(b+c)(a(a+b)+c(a+b))$$
$$(b+c)(a+b)(a+c)$$
$$(a+b)(b+c)(c+a)$$
 *And we are done !!!!!
A: The polynomial in invariant under any permutation of $a,b,c$.
One way to attack this sort of polynomial is express it in terms
of its elementary symmetric polynomials and try to locate/detect any patterns that are useful.
We have $3$ variables, so we have $3$ elementary symmetric polynomials.
Let us call them $A,B,C$. They can be defined using following relations:
$$
(x-a)(x-b)(x-c) = x^3 - Ax^2 + Bx - C
\quad\iff\quad
\begin{cases}
A = a+b+c,\\
B = ab+bc+ca,\\
C = abc
\end{cases}$$
With help of $A,B,C$, we have:
$$\begin{align}
& {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc\\
= & ab(a+b) + bc(b+c)+ ca(c+a) + 2abc\\
= & ab(A-c) + bc(A-a)+ ca(A-b) + 2abc\\
= & (ab+bc+ca)A - abc\\
= & AB-C\\
= & A^3 - A\,A^2 + B\,A - C\\
= & (A-a)(A-b)(A-c)\\
= & (b+c)(c+a)(a+b)
\end{align}
$$
A: The factorization of $ab^2+a^2b+a^2c+ac^2+b^2c+bc^2+2abc$
is $(a + b) (a + c) (b + c)$.
First, take out the $(a+b)$ (because a and b are factors to exactly half of the polynomial):
$$((a^2b+a^2c+ac^2+abc)+(ab^2+b^2c+bc^2+abc))*(\frac{a+b}{a+b})$$Then we have:$$(a+b)(ab+ac+bc+c^2)$$Then take out $(a+c)$:$$(a+b)(a(b+c)+c(b+c))$$And voila! By the commutative property, we have: $$(a+b)(a+c)(b+c)$$
