How to factorize $2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c$?

Question

If $a,b,c$ are in AP $\implies \frac{1}{bc}, \frac{1}{ca}, \frac{1}{ab}$ are in AP $\implies \frac{ab+bc+ca}{bc}, \frac{ab+bc+ca}{ca}, \frac{ab+bc+ca}{ab}$ are in AP $\implies \frac{bc+ca}{bc},\frac{ab+bc}{ca},\frac{bc+ca}{ab}$ are in AP $\implies \frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$ are in HP.

If $P,Q,R$ are in HP $\implies Q=\frac{2PR}{P+R}.$ By this method proving that $\frac{bc}{bc+ca}, \frac{ca}{ab+bc}, \frac{ab}{bc+ca}$, if $a,b,c$ are in AP; is not straight forward.

Curiously, then we need to factorize $(2ab^2+2b^2c-ac^2-bc^2-a^2b-a^2c)$.

The question is: How do you factorize the last expression?

Answer 1

It's tricky! \begin{array}{l} 2ab^2 + 2b^2 c - ac^2 - bc^2 - a^2 b - a^2 c + 2abc- 2abc = \\ \\ = (2ab^2 + 2b^2 c + 2abc) - \left( {a^2 b + a^2 c + abc} \right) - \left( {ac^2 + bc^2 + abc} \right) = \\ \\ = 2b(ab + bc + ac) - a(ab + bc + ac) - c(ab + bc + ac) = \\ \\ = - (ab + bc + ac)\left( {a - 2b + c} \right) \\ \end{array}

Answer 2

If rearranging terms as suggested in the previous answer is difficult to imagine here you have another approach but this has much simplification work.
Let the given polynomial be denoted by f(a,b,c) Since this is homogeneous polynomial of a,b,c when you replace a by x root of x of f(x,b,c) =0 must be in the form of mb+nc. Therefore you can substitute mb+nc for x in f(x,b,c) = 0 and that gives m =2 , n = -1 when you equal the coefficients to zero .
Now x = 2b - c is a root and x - 2b +c is a factor of f(x,b,c) .
Now you can divide f(x,b,c) by x-2b+c to find the other factor and finally x must be changed back to a .

Answer 3

Given expression is $2(a+c)b^2-(a^2+c^2)b-ac(a+c)$

=$2(a+c)b^2-(a+c)^2b+2acb - ac(a+c)=(a+c)b\left[2b-(a+c) \right]+ac\left[2b-(a+c)\right]$

$=(ab+bc+ca)(2a-b-c)$