\(=\dfrac{x^2+x^2a+a+a^2+a^2x^2+1}{x^2-x^2a-a+a^2+a^2x^2+1}\)

\(=\dfrac{x^2\left(a^2+a+1\right)+\left(a^2+a+1\right)}{x^2\left(a^2-a+1\right)+\left(a^2-a+1\right)}\)

\(=\dfrac{\left(x^2+1\right)\left(a^2+a+1\right)}{\left(x^2+1\right)\left(a^2-a+1\right)}\)

\(=\dfrac{a^2+a+1}{a^2-a+1}\)

To \(x^2+1\) other 0

This proves that the given distribution does not depend on x .

Comment form : \(a^2-a+1=\left(a-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\) so the breakdown \(\dfrac{a^2+a+1}{a^2-a+1}\) or \(\dfrac{\left(x^2+a\right)\left(1+a\right)+a^2x^2+1}{\left(x^2-a\right)\left(1-a\right)+a^2x^2+1}\) means for every x and a .