How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?

Question

How can we factorize $a{b^2} + {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc$?

Answer 1

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 !!!!!

Answer 2

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: $$(a+b)(a+c)(b+c)$$

Answer 3

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: $$\begin{align} & {a^2}b + {a^2}c + a{c^2} + {b^2}c + b{c^2} + 2abc\\ = & ab(a+b) + bc(b+c)+ ca(c+a) + 2abc\\ = & ab(A-c) + bc(A-a)+ ca(A-b) + 2abc\\ = & (ab+bc+ca)A - abc\\ = & AB-C\\ = & A^3 - A\,A^2 + B\,A - C\\ = & (A-a)(A-b)(A-c)\\ = & (b+c)(c+a)(a+b) \end{align} $$