If  $\alpha ,\beta$  are two distinct real roots of the equation $a{x}^3+x-1-a=0,(a\ne -1,0)$  none of which is equal to unity, then the value of  $\underset{x\to (1/\alpha )}{\lim }$$\frac{(1+a)x^3-x^2-a}{(e^{1-ax}-1)(x-1)}$  is   $\frac{al(k\alpha -\beta )}{\alpha }$. Find the value of  kl.

$\begin{array}{l}a{x}^3+x-1-a=0\\ \Rightarrow a\left(x^3-1\right)+(x-1)=0\Rightarrow (x-1)\left\{a{x}^2+ax+(a+1)\right\}=0\end{array}$

$\text{ It is given that α.,β are two distinct roots of }a{x}^3+x-1-a=0\text{ or }$

$$\begin{gathered} \text{ Therefore, }\alpha ,\beta \text{ are roots of } \\ a{x}^2+ax+(a+1)=0\text{ and hence, }\frac{1}{\alpha },\frac{1}{\beta }\text{ are roots of } \\ (a+1)x^2+ax+a=0. \\ \therefore (a+1)\left(x-\frac{1}{\alpha }\right)\left(x-\frac{1}{\beta }\right)=(a+1)x^2+ax+a \end{gathered}$$

$\begin{array}{l}\underset{x\to 1/\alpha }{lim} \frac{(1+a)x^3-x^2-a}{\left(e^{1-\alpha x}-1\right)(x-1)}\\ =\underset{x\to 1/\alpha }{lim} \frac{a\left(x^3-1\right)+x^2(x-1)}{\left(e^{1-\alpha x}-1\right)(x-1)}\\ =\underset{x\to 1/\alpha }{lim} \frac{a\left(x^2+x+1\right)+x^2}{\left(e^{1-\alpha x}-1\right)}\\ =\underset{x\to 1/\alpha }{lim} \frac{(a+1)x^2+ax+a}{1-\alpha x}=\\ \\ \left.=\underset{x\to 1/\alpha }{lim} \frac{(a+1)\left(x-\frac{1}{\alpha }\right)(x-{\displaystyle \frac{1}{\beta }})}{1-\alpha x}\right)\\ \\ =-\left(a+1\right)\left(-\frac{1}{\alpha }\right)(\frac{1}{\alpha }-\frac{1}{\beta })\\ \therefore \frac{al}{\alpha }(k\alpha -\beta )=\frac{(a+1)}{{\alpha }^2\beta }(\alpha -\beta )\end{array}$

$\begin{array}{r}\Rightarrow \frac{al}{\alpha }(k\alpha -\beta )=\frac{(a+1)a}{\alpha (a+1)}(\alpha -\beta )\left[\because \alpha \beta =\frac{a+1}{a}\right]\\ \Rightarrow l(k\alpha -\beta )=(\alpha -\beta )\Rightarrow l=k=1\Rightarrow kl=1\end{array}$