MHB Real Factor $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2$

[kaliprasad]

Factor in real $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2$

 

[Theia]

Spoiler

Let $$l = (1-(ab+bc+ca))^2 + (a+b+c-abc)^2$$.

Then by direct simplification one obtains

$$\begin{align*} l &= 1 - 2ab - 2ac - 2bc + a^2b^2 + a^2c^2 + b^2c^2 + 2ab^2c + 2a^2bc + 2abc^2 + a^2 + b^2 + c^2 \\
&\phantom{mmm}+ 2ab + 2ac + 2bc - 2a^2bc - 2ab^2c - 2abc^2 + a^2b^2c^2 \\
&= 1 + a^2b^2 + a^2c^2 + b^2c^2 + a^2 + b^2 + c^2 + a^2b^2c^2 \\
&= 1 + a^2 + a^2(b^2 + c^2) + b^2 + c^2 + b^2c^2(1 + a^2) \\
&= (1 + a^2)(1 + b^2 + c^2 + b^2c^2) \\
&= (1 + a^2)(1 + c^2 + b^2(1 + c^2)) \\
&= (1 + a^2)(1 + b^2)(1 + c^2) \end{align*}$$.

 

[kaliprasad]

My Solution

Spoiler

using $a^2+b^2 = (a+ib)(a-ib)$
we get $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2 = ( 1 +(a+b+c) i - (ab+bc+ca) - abci)) ( 1 - (a+b+c) i - (ab+bc+ca) + abci))$
$= ( 1 +(a+b+c) i + (ab+bc+ca)i^2 + abci^3 )) ( 1 - (a+b+c) i + (ab+bc+ca)^2 - abci^3))$
$= ( 1 +ai)(1+bi)(1+ci)(1- ai)(1-bi)(1-ci)$
$= ( 1 +ai)(1-ai)(a+bi)(1- bi)(1+ci)(1-ci)$
$= ( 1 +a^2)(1+b^2)(1+c^2)$