a)  We have 2 ways to compact :

C1 : \(P=\dfrac{x^6\left(x+1\right)+x^4\left(x+1\right)+x^2\left(x+1\right)+x+1}{x^4-1}\)

\(=\dfrac{\left(x+1\right)\left(x^6+x^4+x^2+1\right)}{\left(x^2-1\right)\left(x^2+1\right)}=\dfrac{\left(x+1\right)\left[x^4\left(x^2+1\right)+x^2+1\right]}{\left(x-1\right)\left(x+1\right)\left(x^2+1\right)}\)

\(=\dfrac{\left(x^2+1\right)\left(x^4+1\right)}{\left(x-1\right)\left(x^2+1\right)}=\dfrac{x^4+1}{x-1}\)

C2 : Multiply numerator and denominator with x-1 we have :

\(P=\dfrac{\left(x-1\right)\left(x^7+x^6+x^5+x^4+x^3+x^2+x+1\right)}{\left(x-1\right)\left(x^4-1\right)}\)

\(=\dfrac{x^8-1}{\left(x-1\right)\left(x^4-1\right)}=\dfrac{\left(x^4-1\right)\left(x^4+1\right)}{\left(x-1\right)\left(x^4-1\right)}=\dfrac{x^4+1}{x-1}\)

b) When x = 2 we have \(P=\dfrac{2^4+1}{2-1}=\dfrac{17}{1}=17\)

So when x = 2 -> P = 17.