Given that ‘a’ is a root of the equation $x^2-x-3=0$ then $\frac{a^3+1}{a^5-a^4-a^3+a^2}$  equals

$\begin{array}{l}{a}^2-a-3=0\\ a^3+1=\left(a+1\right)\left(a^2-a+1\right)\\ a^5-a^4-a^3+a^2=a^2\left(a^3-a^2-a+1\right)\\ =a^2\left(a^2\left(a-1\right)-\left(a-1\right)\right)=a^2\left(a-1\right)\left(a^2-1\right)\\ =\left(a+1\right){\left(a^3-a\right)}^2\\ \text{it follows that }\\ \frac{a^3+1}{a^5-a^4-a^3+a^2}=\frac{a^2-a+1}{{\left(a^2-a\right)}^2}=\frac{4}{3^2}=\frac{4}{9}\end{array}$