Consider the fraction $\frac{x^3-a{x}^2+19x-a-4}{x^3-(a+1)x^2+23x-a-7}$

Subtracting numerator from denominator, we get
$x^2-4x+3$ i.e $(x-1)(x-3)$  
Thus it is concluded that numerator and denominator must be completely divisible by $(x-1)$ or $(x-3)$ in other words both must vanish when $x=1$ or when $x=3,$ if $x=3$ we get, $a=8$  And fraction becomes $\frac{x^3-8x^2+19x-12}{x^3-9x^2+23x-15}=\frac{x^2-7x+12}{x^2-8x+15}=\frac{x-4}{x-5}$
If we put $x=1,$ we get also that $a=8.$