What is the value of $\frac{{\left(x^{(a+b)}\right)}^2{\left(x^{(b+c)}\right)}^2{\left(x^{(c+a)}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}$ after simplifying it?

Given- $\frac{{\left(x^{(a+b)}\right)}^2{\left(x^{(b+c)}\right)}^2{\left(x^{(c+a)}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}$  We can simplify it as,
$\frac{{\left(x^{(a+b)}\right)}^2{\left(x^{(b+c)}\right)}^2{\left(x^{(c+a)}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=\frac{x^{2a+2b+2b+2c+2c+2a}}{x^{4a+4b+4c}}$

$\Rightarrow \frac{{\left(x^{(a+b)}\right)}^2{\left(x^{(b+c)}\right)}^2{\left(x^{(c+a)}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=\frac{x^{4a+4b+4c}}{x^{4a+4b+4c}}$

$\Rightarrow \frac{{\left(x^{(a+b)}\right)}^2{\left(x^{(b+c)}\right)}^2{\left(x^{(c+a)}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}=1$

Thus, the value of $\frac{{\left(x^{(a+b)}\right)}^2{\left(x^{(b+c)}\right)}^2{\left(x^{(c+a)}\right)}^2}{{\left(x^a{x}^b{x}^c\right)}^4}$ after simplifying it is 1.